Trigonometric Equations

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

we have,

We know that sin 30° = sin π/6 = 0.5

∴

∵ it matches with the form sin x = sin y

Hence,

,where nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

we have,

We know that sin 30° = sin π/6 = 0.5

∴

∵ it matches with the form sin x = sin y

Hence,

,where nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that, cos 150° =

∴

If cos x = cos y then x = 2nπ ± y, where n ∈ Z.

For above equation y = 5π / 6

∴ x = 2nπ ± 5π / 6 ,where nϵZ

Thus, x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that, cos 150° =

∴

If cos x = cos y then x = 2nπ ± y, where n ∈ Z.

For above equation y = 5π / 6

∴ x = 2nπ ± 5π / 6 ,where nϵZ

Thus, x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that sin x, and cosec x have negative values in the 3^rd and 4^th quadrant.

While giving a solution, we always try to take the least value of y

The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e., negative angle)

-√2 = -cosec (π/4) = cosec (-π/4) { ∵ sin -θ = -sin θ }

∴

⇒

If sin x = sin y ,then x = nπ + (– 1)ⁿy , where n ∈ Z.

For above equation y =

∴ x = nπ + (-1)ⁿ,where nϵZ

Or x = nπ + (-1)ⁿ⁺¹,where nϵZ

Thus, x gives the required general solution for given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that sin x, and cosec x have negative values in the 3^rd and 4^th quadrant.

While giving a solution, we always try to take the least value of y

The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e., negative angle)

-√2 = -cosec (π/4) = cosec (-π/4) { ∵ sin -θ = -sin θ }

∴

⇒

If sin x = sin y ,then x = nπ + (– 1)ⁿy , where n ∈ Z.

For above equation y =

∴ x = nπ + (-1)ⁿ,where nϵZ

Or x = nπ + (-1)ⁿ⁺¹,where nϵZ

Thus, x gives the required general solution for given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that sec x and cos x have positive values in the 1^st and 4^th quadrant.

While giving a solution, we always try to take the least value of y

both quadrants will give the least magnitude of y.

We can choose any one, in this solution we are assuming a positive value.

⇒

If cos x = cos y then x = 2nπ ± y, where n ∈ Z.

For above equation y = π / 4

∴ x = 2nπ ±,where nϵZ

Thus, x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that sec x and cos x have positive values in the 1^st and 4^th quadrant.

While giving a solution, we always try to take the least value of y

both quadrants will give the least magnitude of y.

We can choose any one, in this solution we are assuming a positive value.

⇒

If cos x = cos y then x = 2nπ ± y, where n ∈ Z.

For above equation y = π / 4

∴ x = 2nπ ±,where nϵZ

Thus, x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that tan x and cot x have negative values in the 2^nd and 4^th quadrant.

While giving solution, we always try to take the least value of y.

The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e. negative angle)

If tan x = tan y then x = nπ + y, where n ∈ Z.

For above equation y =

∴ x = nπ +,w here nϵZ

Or x = nπ -,wh ere nϵZ

Thus, x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that tan x and cot x have negative values in the 2^nd and 4^th quadrant.

While giving solution, we always try to take the least value of y.

The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e. negative angle)

If tan x = tan y then x = nπ + y, where n ∈ Z.

For above equation y =

∴ x = nπ +,w here nϵZ

Or x = nπ -,wh ere nϵZ

Thus, x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

We know that sec x and cos x have positive values in the 1^st and 4^th quadrant.

While giving solution, we always try to take the least value of y

both quadrants will give the least magnitude of y.

We can choose any one, in this solution we are assuming a positive value.

⇒

If cos x = cos y then x = 2nπ ± y, where n ∈ Z.

For above equation y = π / 6

∴ x = 2nπ ±,where nϵZ

Thus, x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

We know that sec x and cos x have positive values in the 1^st and 4^th quadrant.

While giving solution, we always try to take the least value of y

both quadrants will give the least magnitude of y.

We can choose any one, in this solution we are assuming a positive value.

⇒

If cos x = cos y then x = 2nπ ± y, where n ∈ Z.

For above equation y = π / 6

∴ x = 2nπ ±,where nϵZ

Thus, x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that sin x, and cos x have positive values in the 1^st and 2^nd quadrant.

While giving solution, we always try to take the least value of y

The first quadrant will give the least magnitude of y.

∴

If sin x = sin y then x = nπ + (– 1)ⁿ y, where n ∈ Z

Clearly on comparing we have y = π/3

⇒ ,where nϵZ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that sin x, and cos x have positive values in the 1^st and 2^nd quadrant.

While giving solution, we always try to take the least value of y

The first quadrant will give the least magnitude of y.

∴

If sin x = sin y then x = nπ + (– 1)ⁿ y, where n ∈ Z

Clearly on comparing we have y = π/3

⇒ ,where nϵZ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that cos x and sec x have positive values in the 1^st and 4^th quadrant.

While giving solution, we always try to take the least value of y

both quadrant will give the least magnitude of y. We prefer the first quadrant.

∴

If cos x = cos y then x = 2nπ ± y, where n ∈ Z

Clearly on comparing we have y = π/3

⇒ ,where nϵZ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that cos x and sec x have positive values in the 1^st and 4^th quadrant.

While giving solution, we always try to take the least value of y

both quadrant will give the least magnitude of y. We prefer the first quadrant.

∴

If cos x = cos y then x = 2nπ ± y, where n ∈ Z

Clearly on comparing we have y = π/3

⇒ ,where nϵZ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

Using transformation formula:

∴

⇒

∴ cos 5x = 0 or sin 4x = 0

If either of the equation is satisfied, the result will be 0

So we will find the solution individually and then finally combined the solution.

∴ cos 5x = 0

⇒ cos 5x = cos π/2

∴

,where n ϵ Z ………eqn 1

Also,

sin 4x = sin 0

Or ,where n ϵ Z ………eqn 2

From equation 1 and eqn 2,

or,where nϵZ ...ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

Using transformation formula:

∴

⇒

∴ cos 5x = 0 or sin 4x = 0

If either of the equation is satisfied, the result will be 0

So we will find the solution individually and then finally combined the solution.

∴ cos 5x = 0

⇒ cos 5x = cos π/2

∴

,where n ϵ Z ………eqn 1

Also,

sin 4x = sin 0

Or ,where n ϵ Z ………eqn 2

From equation 1 and eqn 2,

or,where nϵZ ...ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒ {∵ sin θ = cos (π/2 – θ) }

If cos x = cos y then x = 2nπ ± y, where n ∈ Z

Clearly on comparing we have y = 3x

∴

, or

∴ or

Hence,

…ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒ {∵ sin θ = cos (π/2 – θ) }

If cos x = cos y then x = 2nπ ± y, where n ∈ Z

Clearly on comparing we have y = 3x

∴

, or

∴ or

Hence,

…ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

We know that: cot θ = tan (π/2 – θ)

∴

⇒ { ∵ - tan θ = tan -θ }

If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒ ,where n ϵ Z …ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

We know that: cot θ = tan (π/2 – θ)

∴

⇒ { ∵ - tan θ = tan -θ }

If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒ ,where n ϵ Z …ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that: cot θ = tan (π/2 – θ)

∴

If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒

∴ ,where nϵZ …..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that: cot θ = tan (π/2 – θ)

∴

If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒

∴ ,where nϵZ …..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

⇒

We know that: cot θ = tan (π/2 – θ)

∴

If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒

⇒ ,where nϵZ ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

⇒

We know that: cot θ = tan (π/2 – θ)

∴

If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒

⇒ ,where nϵZ ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

We know that: cot θ = tan (π/2 – θ)

∴

⇒ { ∵ - tan θ = tan -θ }

If tan x = tan y, then x is given by x = kπ + y, where k ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒

,where k ϵ Z …ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

We know that: cot θ = tan (π/2 – θ)

∴

⇒ { ∵ - tan θ = tan -θ }

If tan x = tan y, then x is given by x = kπ + y, where k ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒

,where k ϵ Z …ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that: cot θ = tan (π/2 – θ)

∴

If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒

∴ ,where nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that: cot θ = tan (π/2 – θ)

∴

If tan x = tan y, then x is given by x = nπ + y, where n ∈ Z.

From above expression, on comparison with standard equation we have

y =

∴

⇒

∴ ,where nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that: sin θ = cos (π/2 – θ)

∴

⇒

We know that: -cos θ = cos (π – θ)

∴

⇒

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y =

∴

Hence,

or

∴ or

⇒ or

∴,where nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that: sin θ = cos (π/2 – θ)

∴

⇒

We know that: -cos θ = cos (π – θ)

∴

⇒

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y =

∴

Hence,

or

∴ or

⇒ or

∴,where nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

⇒

⇒

either,

sin x = 0 or cos x = 1

⇒ sin x = sin 0 or cos x = cos 0

We know that,

If sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z

∵ sin x = sin 0

∴ y = 0

And hence,

x = nπ where nϵZ

Also,

If cos x = cos y, implies x = 2mπ ±y, where m ∈ Z

∵ cos x = cos 0

∴ y = 0

Hence, x is given by

x = 2mπ where mϵZ

∴x = nπ or 2mπ ,where m,nϵZ …ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

⇒

⇒

either,

sin x = 0 or cos x = 1

⇒ sin x = sin 0 or cos x = cos 0

We know that,

If sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z

∵ sin x = sin 0

∴ y = 0

And hence,

x = nπ where nϵZ

Also,

If cos x = cos y, implies x = 2mπ ±y, where m ∈ Z

∵ cos x = cos 0

∴ y = 0

Hence, x is given by

x = 2mπ where mϵZ

∴x = nπ or 2mπ ,where m,nϵZ …ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that: sin θ = cos (π/2 – θ)

∴

⇒

We know that: -cos θ = cos (π – θ)

∴

⇒

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y =

∴

Hence,

or

∴ or

⇒ or

∴,where nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

We know that: sin θ = cos (π/2 – θ)

∴

⇒

We know that: -cos θ = cos (π – θ)

∴

⇒

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y =

∴

Hence,

or

∴ or

⇒ or

∴,where nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

As, sin² x = 1 – cos² x

∴ we have,

⇒

⇒

Let, cos x = k

∴ 4k² + 4k – 3 = 0

⇒ 4k² -2k + 6k – 3

⇒ 2k(2k – 1) +3(2k – 1) = 0

⇒ (2k – 1)(2k + 3) = 0

∴ k =1/2 or k = -3/2

⇒ cos x = � or cos x = -3/2

As cos x lies between -1 and 1

∴ cos x can’t be -3/2

So we ignore that value.

∴ cos x = �

⇒ cos x = cos 60° = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

On comparing our equation with standard form, we have

y = π/3

∴ where nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

As, sin² x = 1 – cos² x

∴ we have,

⇒

⇒

Let, cos x = k

∴ 4k² + 4k – 3 = 0

⇒ 4k² -2k + 6k – 3

⇒ 2k(2k – 1) +3(2k – 1) = 0

⇒ (2k – 1)(2k + 3) = 0

∴ k =1/2 or k = -3/2

⇒ cos x = � or cos x = -3/2

As cos x lies between -1 and 1

∴ cos x can’t be -3/2

So we ignore that value.

∴ cos x = �

⇒ cos x = cos 60° = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

On comparing our equation with standard form, we have

y = π/3

∴ where nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

2 cos² x – 5 cos x + 2 =0

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

Let, cos x = k

∴ 2k² – 5k + 2 = 0

⇒ 2k² – 4k – k +2 = 0

⇒ 2k(k – 2) -1(k -2) = 0

⇒ (k – 2)(2k - 1) = 0

∴ k = 2 or k = �

⇒ cos x = 2 {which is not possible} or cos x = � (acceptable)

∴ cos x = �

⇒ cos x = cos 60° = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

On comparing our equation with standard form, we have

y = π/3

∴ where nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

2 cos² x – 5 cos x + 2 =0

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

Let, cos x = k

∴ 2k² – 5k + 2 = 0

⇒ 2k² – 4k – k +2 = 0

⇒ 2k(k – 2) -1(k -2) = 0

⇒ (k – 2)(2k - 1) = 0

∴ k = 2 or k = �

⇒ cos x = 2 {which is not possible} or cos x = � (acceptable)

∴ cos x = �

⇒ cos x = cos 60° = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

On comparing our equation with standard form, we have

y = π/3

∴ where nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

2sin² x +√3 cos x + 1 = 0

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

As, sin² x = 1 – cos² x

∴ we have,

⇒

⇒

Let, cos x = k

∴ 2k² - √3 k – 3 = 0

⇒ 2k² -2√3 k + √3 k – 3 = 0

⇒ 2k(k – √3) +√3(k – √3) = 0

⇒ (2k + √3)(k - √3) = 0

∴ k = √3 or k = -√3/2

⇒ cos x = √3 or cos x = -√3/2

As cos x lies between -1 and 1

∴ cos x can’t be √3

So we ignore that value.

∴ cos x = -√3/2

⇒ cos x = cos 150° = cos 5π/6

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

On comparing our equation with standard form, we have

y = 5π/6

∴ where nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

2sin² x +√3 cos x + 1 = 0

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

As, sin² x = 1 – cos² x

∴ we have,

⇒

⇒

Let, cos x = k

∴ 2k² - √3 k – 3 = 0

⇒ 2k² -2√3 k + √3 k – 3 = 0

⇒ 2k(k – √3) +√3(k – √3) = 0

⇒ (2k + √3)(k - √3) = 0

∴ k = √3 or k = -√3/2

⇒ cos x = √3 or cos x = -√3/2

As cos x lies between -1 and 1

∴ cos x can’t be √3

So we ignore that value.

∴ cos x = -√3/2

⇒ cos x = cos 150° = cos 5π/6

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

On comparing our equation with standard form, we have

y = 5π/6

∴ where nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

4sin² x -8 cos x + 1 = 0

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

As, sin² x = 1 – cos² x

∴ we have,

⇒

⇒

Let, cos x = k

∴ 4k² + 8k – 5 = 0

⇒ 4k² -2k + 10k – 5 = 0

⇒ 2k(2k – 1) +5(2k – 1) = 0

⇒ (2k + 5)(2k - 1) = 0

∴ k = -5/2 = -2.5 or k = 1/2

⇒ cos x = -2.5 or cos x = 1/2

As cos x lies between -1 and 1

∴ cos x can’t be -2.5

So we ignore that value.

∴ cos x = 1/2

⇒ cos x = cos 60° = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

On comparing our equation with standard form, we have

y = π/3

∴ where nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

4sin² x -8 cos x + 1 = 0

As the equation is of 2^nd degree, so we need to solve a quadratic equation.

First we will substitute trigonometric ratio with some variable k and we will solve for k

As, sin² x = 1 – cos² x

∴ we have,

⇒

⇒

Let, cos x = k

∴ 4k² + 8k – 5 = 0

⇒ 4k² -2k + 10k – 5 = 0

⇒ 2k(2k – 1) +5(2k – 1) = 0

⇒ (2k + 5)(2k - 1) = 0

∴ k = -5/2 = -2.5 or k = 1/2

⇒ cos x = -2.5 or cos x = 1/2

As cos x lies between -1 and 1

∴ cos x can’t be -2.5

So we ignore that value.

∴ cos x = 1/2

⇒ cos x = cos 60° = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

On comparing our equation with standard form, we have

y = π/3

∴ where nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

⇒

⇒

∴ tan x = -1 or tan x = √3

As, tan x ϵ (-∞ , ∞) so both values are valid and acceptable.

⇒ tan x = tan (-π/4) or tan x = tan (π/3)

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

Clearly by comparing standard form with obtained equation we have

y = -π/4 or y = π/3

∴ or

Hence,

,where m,nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

⇒

⇒

⇒

∴ tan x = -1 or tan x = √3

As, tan x ϵ (-∞ , ∞) so both values are valid and acceptable.

⇒ tan x = tan (-π/4) or tan x = tan (π/3)

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

Clearly by comparing standard form with obtained equation we have

y = -π/4 or y = π/3

∴ or

Hence,

,where m,nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

⇒

⇒

⇒

∴ either, or

⇒ or

⇒ or

⇒ or

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

Clearly by comparing standard form with obtained equation we have:

y = π/6 or y = -π/3

∴ or

Hence,

,where m,nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

⇒

⇒

⇒

∴ either, or

⇒ or

⇒ or

⇒ or

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

Clearly by comparing standard form with obtained equation we have:

y = π/6 or y = -π/3

∴ or

Hence,

,where m,nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y = 2x

∴ 4

Hence,

or

∴ or

⇒ x = nπ or

∴where m, nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y = 2x

∴ 4

Hence,

or

∴ or

⇒ x = nπ or

∴where m, nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x + cos 2x + cos 3x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, cos x + cos 2x + cos 3x = 0

∴ we will use cos x and cos 2x for transformation as after transformation it will give cos 2x term which can be taken common.

∴ cos x + cos 2x + cos 3x = 0

⇒ cos 2x + (cos x + cos 3x) = 0

{∵ cos A + cos B = 2

⇒ cos 2x + 2 cos

⇒ cos 2x + 2cos 2x cos x = 0

⇒ cos 2x ( 1 + 2 cos x) = 0

∴ cos 2x = 0 or 1 + 2cos x = 0

⇒ cos 2x = cos π/2 or cos x = -1/2

⇒ cos 2x = cos π/2 or cos x = cos (π - π/3) = cos (2π /3)

If cos x = cos y implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y = π/2 or y = 2π/3

∴ 2x = 2nπ ± π/2 or x = 2mπ ± 2π/3

∴ where m, nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x + cos 2x + cos 3x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, cos x + cos 2x + cos 3x = 0

∴ we will use cos x and cos 2x for transformation as after transformation it will give cos 2x term which can be taken common.

∴ cos x + cos 2x + cos 3x = 0

⇒ cos 2x + (cos x + cos 3x) = 0

{∵ cos A + cos B = 2

⇒ cos 2x + 2 cos

⇒ cos 2x + 2cos 2x cos x = 0

⇒ cos 2x ( 1 + 2 cos x) = 0

∴ cos 2x = 0 or 1 + 2cos x = 0

⇒ cos 2x = cos π/2 or cos x = -1/2

⇒ cos 2x = cos π/2 or cos x = cos (π - π/3) = cos (2π /3)

If cos x = cos y implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y = π/2 or y = 2π/3

∴ 2x = 2nπ ± π/2 or x = 2mπ ± 2π/3

∴ where m, nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x - cos 2x + cos 3x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, cos x - cos 2x + cos 3x = 0

∴ we will use cos x and cos 2x for transformation as after transformation it will give cos 2x term which can be taken common.

∴ cos x - cos 2x + cos 3x = 0

⇒ -cos 2x + (cos x + cos 3x) = 0

{∵ cos A + cos B = 2

⇒ -cos 2x + 2 cos

⇒ -cos 2x + 2cos 2x cos x = 0

⇒ cos 2x ( -1 + 2 cos x) = 0

∴ cos 2x = 0 or 1 + 2cos x = 0

⇒ cos 2x = cos π/2 or cos x = 1/2

⇒ cos 2x = cos π/2 or cos x = cos (π/3) = cos (π /3)

If cos x = cos y implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y = π/2 or y = π/3

∴ 2x = 2nπ ± π/2 or x = 2mπ ± π/3

∴where m, nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x - cos 2x + cos 3x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, cos x - cos 2x + cos 3x = 0

∴ we will use cos x and cos 2x for transformation as after transformation it will give cos 2x term which can be taken common.

∴ cos x - cos 2x + cos 3x = 0

⇒ -cos 2x + (cos x + cos 3x) = 0

{∵ cos A + cos B = 2

⇒ -cos 2x + 2 cos

⇒ -cos 2x + 2cos 2x cos x = 0

⇒ cos 2x ( -1 + 2 cos x) = 0

∴ cos 2x = 0 or 1 + 2cos x = 0

⇒ cos 2x = cos π/2 or cos x = 1/2

⇒ cos 2x = cos π/2 or cos x = cos (π/3) = cos (π /3)

If cos x = cos y implies x = 2nπ ± y, where n ∈ Z.

From above expression and on comparison with standard equation we have:

y = π/2 or y = π/3

∴ 2x = 2nπ ± π/2 or x = 2mπ ± π/3

∴where m, nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin x + sin 5x = sin 3x

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, sin x + sin 5x = sin 3x

∴ sin x + sin 5x – sin 3x = 0

∴ we will use sin x and sin 5x for transformation as after transformation it will give sin 3x term which can be taken common.

{∵ sin A + sin B =

⇒ -sin 3x + 2 sin

⇒ 2sin 3x cos 2x – sin 3x = 0

⇒ sin 3x ( 2cos 2x – 1) = 0

∴ either, sin 3x = 0 or 2cos 2x – 1 = 0

⇒ sin 3x = sin 0 or cos 2x = � = cos π/3

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

Comparing obtained equation with standard equation, we have:

3x = nπ or 2x = 2mπ ± π/3

∴ where m,nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin x + sin 5x = sin 3x

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, sin x + sin 5x = sin 3x

∴ sin x + sin 5x – sin 3x = 0

∴ we will use sin x and sin 5x for transformation as after transformation it will give sin 3x term which can be taken common.

{∵ sin A + sin B =

⇒ -sin 3x + 2 sin

⇒ 2sin 3x cos 2x – sin 3x = 0

⇒ sin 3x ( 2cos 2x – 1) = 0

∴ either, sin 3x = 0 or 2cos 2x – 1 = 0

⇒ sin 3x = sin 0 or cos 2x = � = cos π/3

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

Comparing obtained equation with standard equation, we have:

3x = nπ or 2x = 2mπ ± π/3

∴ where m,nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x cos 2x cos 3x = �

⇒ 4cos x cos 2x cos 3x – 1 = 0

{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}

∴ 2(2cos x cos 3x)cos 2x – 1 = 0

⇒ 2(cos 4x + cos 2x)cos2x – 1 = 0

⇒ 2(2cos² 2x – 1 + cos 2x)cos 2x – 1 = 0 {using cos 2θ = 2cos²θ – 1 }

⇒ 4cos³ 2x – 2cos 2x + 2cos² 2x – 1 = 0

⇒ 2cos² 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0

⇒ (2cos² 2x – 1)(2 cos 2x + 1) = 0

∴ either, 2cos 2x + 1 = 0 or (2cos² 2x – 1) = 0

⇒ cos 2x = -1/2 or cos 4x = 0 {using cos 2θ = 2cos²θ – 1}

⇒ cos 2x = cos (π - π/3) = cos 2π /3 or cos 4x = cos π/2

If cos x = cos y implies x = 2nπ ± y, where n ∈ Z.

In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2

Comparing obtained equation with standard equation, we have:

y = 2π / 3 or y = π/2

∴ 2x = 2mπ ± 2π/3 or 4x = (2n+1)π/2

∴where m,nϵZ ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x cos 2x cos 3x = �

⇒ 4cos x cos 2x cos 3x – 1 = 0

{∵ 2 cos A cos B = cos (A + B) + cos (A – B)}

∴ 2(2cos x cos 3x)cos 2x – 1 = 0

⇒ 2(cos 4x + cos 2x)cos2x – 1 = 0

⇒ 2(2cos² 2x – 1 + cos 2x)cos 2x – 1 = 0 {using cos 2θ = 2cos²θ – 1 }

⇒ 4cos³ 2x – 2cos 2x + 2cos² 2x – 1 = 0

⇒ 2cos² 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0

⇒ (2cos² 2x – 1)(2 cos 2x + 1) = 0

∴ either, 2cos 2x + 1 = 0 or (2cos² 2x – 1) = 0

⇒ cos 2x = -1/2 or cos 4x = 0 {using cos 2θ = 2cos²θ – 1}

⇒ cos 2x = cos (π - π/3) = cos 2π /3 or cos 4x = cos π/2

If cos x = cos y implies x = 2nπ ± y, where n ∈ Z.

In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2

Comparing obtained equation with standard equation, we have:

y = 2π / 3 or y = π/2

∴ 2x = 2mπ ± 2π/3 or 4x = (2n+1)π/2

∴where m,nϵZ ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x + sin x = cos 2x + sin 2x

cos x – cos 2x = sin 2x – sin x

{∵ sin A - sin B =

∴

⇒

∴

Hence,

Either,

⇒

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴

⇒ where m,n ϵ Z ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x + sin x = cos 2x + sin 2x

cos x – cos 2x = sin 2x – sin x

{∵ sin A - sin B =

∴

⇒

∴

Hence,

Either,

⇒

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴

⇒ where m,n ϵ Z ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin x + sin 2x + sin 3x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, sin x + sin 2x + sin 3x = 0

∴ we will use sin x and sin 3x for transformation as after transformation it will give sin 2x term which can be taken common.

{∵ sin A + sin B =

⇒ sin 2x + 2 sin

⇒ 2sin 2x cos x + sin 2x = 0

⇒ sin 2x ( 2cos x + 1) = 0

∴ either, sin 2x = 0 or 2cos x + 1 = 0

⇒ sin 2x = sin 0 or cos x = - � = cos (π-π/3) = cos 2π/3

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

Comparing obtained equation with standard equation, we have:

2x = nπ or x = 2mπ ± 2π/3

∴ where m,nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin x + sin 2x + sin 3x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, sin x + sin 2x + sin 3x = 0

∴ we will use sin x and sin 3x for transformation as after transformation it will give sin 2x term which can be taken common.

{∵ sin A + sin B =

⇒ sin 2x + 2 sin

⇒ 2sin 2x cos x + sin 2x = 0

⇒ sin 2x ( 2cos x + 1) = 0

∴ either, sin 2x = 0 or 2cos x + 1 = 0

⇒ sin 2x = sin 0 or cos x = - � = cos (π-π/3) = cos 2π/3

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

Comparing obtained equation with standard equation, we have:

2x = nπ or x = 2mπ ± 2π/3

∴ where m,nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin x + sin 2x + sin 3x + sin 4x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, sin x + sin 2x + sin 3x + sin 4x= 0

∴ we will use sin x and sin 3x together in 1 group for transformation and sin 4x and sin 2x common in other group as after transformation both will give cos x term which can be taken common.

{∵ sin A + sin B =

(sin x + sin 3x )+( sin 2x + sin 4x) = 0

⇒ 2 sin + 2 sin

⇒ 2sin 2x cos x + 2sin 3x cos x= 0

⇒ 2cos x (sin 2x + sin 3x) = 0

Again using transformation formula, we have:

⇒ 2 cos x 2

⇒

∴ either, cos x = 0 or or

In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2

In case of sin x = 0 we can give solution directly as sin x = 0 is true for x = integral multiple of π

∴

⇒ where n,p,mϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin x + sin 2x + sin 3x + sin 4x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

As, sin x + sin 2x + sin 3x + sin 4x= 0

∴ we will use sin x and sin 3x together in 1 group for transformation and sin 4x and sin 2x common in other group as after transformation both will give cos x term which can be taken common.

{∵ sin A + sin B =

(sin x + sin 3x )+( sin 2x + sin 4x) = 0

⇒ 2 sin + 2 sin

⇒ 2sin 2x cos x + 2sin 3x cos x= 0

⇒ 2cos x (sin 2x + sin 3x) = 0

Again using transformation formula, we have:

⇒ 2 cos x 2

⇒

∴ either, cos x = 0 or or

In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2

In case of sin x = 0 we can give solution directly as sin x = 0 is true for x = integral multiple of π

∴

⇒ where n,p,mϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin 3x – sin x = 4 cos² x – 2

⇒ sin 3x – sin x = 2(2 cos² x – 1)

⇒ sin 3x – sin x = 2 cos 2x {∵ cos 2θ = 2cos² θ – 1}

{∵ sin A - sin B =

⇒

⇒

⇒

∴ either, cos 2x = 0 or sin x = 1 = sin π/2

In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2

If sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

∴

⇒ where m, nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin 3x – sin x = 4 cos² x – 2

⇒ sin 3x – sin x = 2(2 cos² x – 1)

⇒ sin 3x – sin x = 2 cos 2x {∵ cos 2θ = 2cos² θ – 1}

{∵ sin A - sin B =

⇒

⇒

⇒

∴ either, cos 2x = 0 or sin x = 1 = sin π/2

In case of cos x = 0 we can give solution directly as cos x = 0 is true for x = odd multiple of π/2

If sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

∴

⇒ where m, nϵZ

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin 2x - sin 4x + sin 6x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

we have, sin 2x - sin 4x + sin 6x = 0

∴ we will use sin 6x and sin 2x for transformation as after transformation it will give sin 4x term which can be taken common.

{∵ sin A + sin B =

⇒ -sin 4x + 2 sin

⇒ 2sin 4x cos 2x – sin 4x = 0

⇒ sin 4x ( 2cos 2x – 1) = 0

∴ either, sin 4x = 0 or 2cos 2x – 1 = 0

⇒ sin 4x = sin 0 or cos 2x = � = cos π/3

If sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

Comparing obtained equation with standard equation, we have:

4x = nπ or 2x = 2mπ ± π/3

∴ where m,nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

sin 2x - sin 4x + sin 6x = 0

To solve the equation we need to change its form so that we can equate the t-ratios individually.

For this we will be applying transformation formulae. While applying the

Transformation formula we need to select the terms wisely which we want

to transform.

we have, sin 2x - sin 4x + sin 6x = 0

∴ we will use sin 6x and sin 2x for transformation as after transformation it will give sin 4x term which can be taken common.

{∵ sin A + sin B =

⇒ -sin 4x + 2 sin

⇒ 2sin 4x cos 2x – sin 4x = 0

⇒ sin 4x ( 2cos 2x – 1) = 0

∴ either, sin 4x = 0 or 2cos 2x – 1 = 0

⇒ sin 4x = sin 0 or cos 2x = � = cos π/3

If sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

Comparing obtained equation with standard equation, we have:

4x = nπ or 2x = 2mπ ± π/3

∴ where m,nϵZ ..ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

tan x + tan 2x + tan 3x = 0

In order to solve the equation we need to reduce the equation into factor form so that we can equate the ratios with 0 and can solve the equation easily

As if we expand tan 3x = tan ( x + 2x) we will get tan x + tan 2x common.

∴ tan x + tan 2x + tan 3x = 0

⇒ tan x + tan 2x + tan (x + 2x) = 0

As, tan (A + B) =

∴ tan x + tan 2x +

⇒

⇒

∴ tan x + tan 2x = 0 or 2 – tan x tan 2x = 0

Using, tan 2x = we have,

⇒ tan x = tan (-2x) or 2 –

⇒ tan x = tan(-2x) or 2 – 4tan² x = 0 ⇒ tan x = 1/ √2

Let 1/√2 = tan α and if tan x = tan y, implies x = nπ + y, where n ∈ Z

∴ x = nπ + (-2x) or tan x = tan α ⇒ x = mπ + α

⇒ 3x = nπ or x = mπ + α

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

tan x + tan 2x + tan 3x = 0

In order to solve the equation we need to reduce the equation into factor form so that we can equate the ratios with 0 and can solve the equation easily

As if we expand tan 3x = tan ( x + 2x) we will get tan x + tan 2x common.

∴ tan x + tan 2x + tan 3x = 0

⇒ tan x + tan 2x + tan (x + 2x) = 0

As, tan (A + B) =

∴ tan x + tan 2x +

⇒

⇒

∴ tan x + tan 2x = 0 or 2 – tan x tan 2x = 0

Using, tan 2x = we have,

⇒ tan x = tan (-2x) or 2 –

⇒ tan x = tan(-2x) or 2 – 4tan² x = 0 ⇒ tan x = 1/ √2

Let 1/√2 = tan α and if tan x = tan y, implies x = nπ + y, where n ∈ Z

∴ x = nπ + (-2x) or tan x = tan α ⇒ x = mπ + α

⇒ 3x = nπ or x = mπ + α

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

tan x + tan 2x - tan 3x = 0

In order to solve the equation we need to reduce the equation into factor form so that we can equate the ratios with 0 and can solve the equation easily

As if we expand tan 3x = tan ( x + 2x) we will get tan x + tan 2x common.

∴ tan x + tan 2x - tan 3x = 0

⇒ tan x + tan 2x - tan (x + 2x) = 0

As, tan (A + B) =

∴ tan x + tan 2x -

⇒

⇒

∴ tan x + tan 2x = 0 or – tan x tan 2x = 0

Using, tan 2x = we have,

⇒ tan x = tan (-2x) or

⇒ tan x = tan(-2x) or tan x = 0 = tan 0

if tan x = tan y, implies x = nπ + y, where n ∈ Z

∴ x = nπ + (-2x) or x = mπ + 0

⇒ 3x = nπ or x = mπ

∴ ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

tan x + tan 2x - tan 3x = 0

In order to solve the equation we need to reduce the equation into factor form so that we can equate the ratios with 0 and can solve the equation easily

As if we expand tan 3x = tan ( x + 2x) we will get tan x + tan 2x common.

∴ tan x + tan 2x - tan 3x = 0

⇒ tan x + tan 2x - tan (x + 2x) = 0

As, tan (A + B) =

∴ tan x + tan 2x -

⇒

⇒

∴ tan x + tan 2x = 0 or – tan x tan 2x = 0

Using, tan 2x = we have,

⇒ tan x = tan (-2x) or

⇒ tan x = tan(-2x) or tan x = 0 = tan 0

if tan x = tan y, implies x = nπ + y, where n ∈ Z

∴ x = nπ + (-2x) or x = mπ + 0

⇒ 3x = nπ or x = mπ

∴ ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

tan x + tan 3x = 2tan 2x

⇒ tan x + tan 3x = tan 2x + tan 2x

⇒ tan 3x – tan 2x = tan 2x – tan x

⇒

As, tan (A - B) =

∴

⇒

⇒

∴ tan x = 0 or tan 2x = 0 or tan 3x = tan x

if tan x = tan y, implies x = nπ + y, where n ∈ Z

∴ x = nπ or 2x = mπ or 3x = kπ + x

∴

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

tan x + tan 3x = 2tan 2x

⇒ tan x + tan 3x = tan 2x + tan 2x

⇒ tan 3x – tan 2x = tan 2x – tan x

⇒

As, tan (A - B) =

∴

⇒

⇒

∴ tan x = 0 or tan 2x = 0 or tan 3x = tan x

if tan x = tan y, implies x = nπ + y, where n ∈ Z

∴ x = nπ or 2x = mπ or 3x = kπ + x

∴

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

In all such problems we try to reduce the equation in an equation involving single trigonometric expression.

∴

⇒ { ∵ }

⇒ { ∵ sin A cos B + cos A sin B = sin (A +B)}

⇒

NOTE: We can also make the ratio of cos instead of sin, the answer remains same but the form of answer may look different, when you put values of n you will get same values with both forms

If sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z

∴

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

In all such problems we try to reduce the equation in an equation involving single trigonometric expression.

∴

⇒ { ∵ }

⇒ { ∵ sin A cos B + cos A sin B = sin (A +B)}

⇒

NOTE: We can also make the ratio of cos instead of sin, the answer remains same but the form of answer may look different, when you put values of n you will get same values with both forms

If sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z

∴

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

In all such problems we try to reduce the equation in an equation involving single trigonometric expression.

∴

⇒ { ∵ }

⇒ { ∵ cos A cos B + sin A sin B = cos (A - B)}

⇒

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

∴

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

In all such problems we try to reduce the equation in an equation involving single trigonometric expression.

∴

⇒ { ∵ }

⇒ { ∵ cos A cos B + sin A sin B = cos (A - B)}

⇒

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

∴

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

.

In all such problems we try to reduce the equation in an equation involving single trigonometric expression.

∴ { dividing by √2 both sides}

⇒ { ∵ }

⇒ { ∵ cos A cos B + sin A sin B = cos (A - B)}

NOTE: We can also make the ratio of sin instead of cos , the answer remains same but the form of answer may look different, when you put values of n you will get same values with both forms

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

∴ s

∴ .

.

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

.

In all such problems we try to reduce the equation in an equation involving single trigonometric expression.

∴ { dividing by √2 both sides}

⇒ { ∵ }

⇒ { ∵ cos A cos B + sin A sin B = cos (A - B)}

NOTE: We can also make the ratio of sin instead of cos , the answer remains same but the form of answer may look different, when you put values of n you will get same values with both forms

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

∴ s

∴ .

.

deas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

cosec x = 1 + cot x

⇒

⇒ .

In all such problems we try to reduce the equation in an equation involving single trigonometric expression.

∴ s { dividing by √2 both sides}

⇒ . { ∵ . }

⇒ . { ∵ cos A cos B + sin A sin B = cos (A - B)}

NE: We can also make the ratio of sin instead of cos , the answer remains same but the form of answer may look different, when you put values of n you will get same values with both forms

If cos x = cos y, impls x = 2nπ ± y, where n ∈ Z

∴ .

∴

deas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

cosec x = 1 + cot x

⇒

⇒ .

In all such problems we try to reduce the equation in an equation involving single trigonometric expression.

∴ s { dividing by √2 both sides}

⇒ . { ∵ . }

⇒ . { ∵ cos A cos B + sin A sin B = cos (A - B)}

NE: We can also make the ratio of sin instead of cos , the answer remains same but the form of answer may look different, when you put values of n you will get same values with both forms

If cos x = cos y, impls x = 2nπ ± y, where n ∈ Z

∴ .

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

Dividing both sides by 2√2 :

We have,

.

⇒ where cos α = π /4

⇒ { cos π/4 = 1/√2 }

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

Dividing both sides by 2√2 :

We have,

.

⇒ where cos α = π /4

⇒ { cos π/4 = 1/√2 }

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

cot x + tan x = 2

⇒

⇒ ( tan x – 1 )² = 0

∴ tan x = 1 ⇒ tan x = tan π/4

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴x = nπ + π/4 where nϵZ ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

cot x + tan x = 2

⇒

⇒ ( tan x – 1 )² = 0

∴ tan x = 1 ⇒ tan x = tan π/4

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴x = nπ + π/4 where nϵZ ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

2 sin² x = 3 cos x , 0 ≤ x ≤ 2π

⇒ 2 ( 1 – cos² x) = 3 cos x

⇒ 2 cos² x + 3cos x – 2 = 0

⇒ 2 cos² x + 4 cos x – cos x – 2 = 0

⇒ 2 cos x(cos x + 2) – 1(cos x + 2) = 0

⇒ (2cos x – 1)(cos x + 2) = 0

∴ cos x = �

or cos x = -2 { as cos x lies between -1 and 1 so this value is rejected }

∴ cos x = � = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

∴ x = 2nπ ± π/3

But, 0 ≤ x ≤ 2π

∴ x = π/3 and x = 2π - π/3 = 5π/3 ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

2 sin² x = 3 cos x , 0 ≤ x ≤ 2π

⇒ 2 ( 1 – cos² x) = 3 cos x

⇒ 2 cos² x + 3cos x – 2 = 0

⇒ 2 cos² x + 4 cos x – cos x – 2 = 0

⇒ 2 cos x(cos x + 2) – 1(cos x + 2) = 0

⇒ (2cos x – 1)(cos x + 2) = 0

∴ cos x = �

or cos x = -2 { as cos x lies between -1 and 1 so this value is rejected }

∴ cos x = � = cos π/3

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z

∴ x = 2nπ ± π/3

But, 0 ≤ x ≤ 2π

∴ x = π/3 and x = 2π - π/3 = 5π/3 ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

sec x cos 5x + 1 = 0, 0 < x < π/2

⇒ sec x cos 5x = -1

⇒ cos 5x = - cos x

∵ - cos x = cos (π – x)

∴ cos 5x = cos (π – x)

If cos x = cos y, implies 2nπ ± y, where n ∈ Z.

∴ 5x = 2nπ ± (π – x)

⇒ 5x = 2nπ + (π – x) or 5x = 2nπ – (π – x)

⇒ 6x = (2n+1)π or 4x = (2n-1)π

∴

But, 0 < x < π/2

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

sec x cos 5x + 1 = 0, 0 < x < π/2

⇒ sec x cos 5x = -1

⇒ cos 5x = - cos x

∵ - cos x = cos (π – x)

∴ cos 5x = cos (π – x)

If cos x = cos y, implies 2nπ ± y, where n ∈ Z.

∴ 5x = 2nπ ± (π – x)

⇒ 5x = 2nπ + (π – x) or 5x = 2nπ – (π – x)

⇒ 6x = (2n+1)π or 4x = (2n-1)π

∴

But, 0 < x < π/2

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

5 cos² x + 7 sin² x – 6 = 0

⇒ 5 cos² x + 5 sin² x + 2sin² x – 6 = 0

⇒ 2 sin² x – 6 + 5 = 0 {∵ sin² x + cos² x = 1}

⇒ 2 sin² x – 1 = 0

⇒ sin² x = (1/2)

∴ sin x = ±(1 /√2)

⇒ sin x = ± sin π /4

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

∴ x = nπ + (-1)ⁿ (±(π / 4)) where n ϵ Z

∴….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

5 cos² x + 7 sin² x – 6 = 0

⇒ 5 cos² x + 5 sin² x + 2sin² x – 6 = 0

⇒ 2 sin² x – 6 + 5 = 0 {∵ sin² x + cos² x = 1}

⇒ 2 sin² x – 1 = 0

⇒ sin² x = (1/2)

∴ sin x = ±(1 /√2)

⇒ sin x = ± sin π /4

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

∴ x = nπ + (-1)ⁿ (±(π / 4)) where n ϵ Z

∴….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x

⇒ (sin x + sin 3x) – 3sin 2x – (cos x + cos 3x) + 3 cox 2x = 0

∵ sin A + sin B =

∴

⇒ 2 sin 2x cos x – 3 sin 2x – 2 cos 2x cos x + 3 cos 2x = 0

⇒ sin 2x ( 2cos x – 3) - cos 2x (2cos x – 3) = 0

⇒ (2cos x – 3)(sin 2x – cos 2x) = 0

∴ cos x = 3/2 = 1.5 (not accepted as cos x lies between – 1 and 1)

Or sin 2x = cos 2x

∴ tan 2x = 1 = tan π/4

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴ 2x = nπ + π/4

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x

⇒ (sin x + sin 3x) – 3sin 2x – (cos x + cos 3x) + 3 cox 2x = 0

∵ sin A + sin B =

∴

⇒ 2 sin 2x cos x – 3 sin 2x – 2 cos 2x cos x + 3 cos 2x = 0

⇒ sin 2x ( 2cos x – 3) - cos 2x (2cos x – 3) = 0

⇒ (2cos x – 3)(sin 2x – cos 2x) = 0

∴ cos x = 3/2 = 1.5 (not accepted as cos x lies between – 1 and 1)

Or sin 2x = cos 2x

∴ tan 2x = 1 = tan π/4

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴ 2x = nπ + π/4

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

4 sin x cos x + 2 sin x + 2 cos x + 1 = 0

⇒ 2sin x (2cos x + 1) + 1(2cos x + 1) = 0

⇒ (2cos x + 1)(2sin x + 1) = 0

∴ cos x = -1/2 or sin x = -1/2

⇒ cos x = cos (π - π/3) or sin x = sin (- π/6)

⇒ cos x = cos 2π/3 or sin x = sin (-π/6)

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

And cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

∴ x = 2nπ ± 2π/3 or x = mπ + (-1)^m (-π/6)

Hence,

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

4 sin x cos x + 2 sin x + 2 cos x + 1 = 0

⇒ 2sin x (2cos x + 1) + 1(2cos x + 1) = 0

⇒ (2cos x + 1)(2sin x + 1) = 0

∴ cos x = -1/2 or sin x = -1/2

⇒ cos x = cos (π - π/3) or sin x = sin (- π/6)

⇒ cos x = cos 2π/3 or sin x = sin (-π/6)

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

And cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

∴ x = 2nπ ± 2π/3 or x = mπ + (-1)^m (-π/6)

Hence,

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x + sin x = cos 2x + sin 2x

cos x – cos 2x = sin 2x – sin x

{∵ sin A - sin B =

∴

⇒ .

∴ .

Hence,

Either,

⇒

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴

⇒ where m,n ϵ Z ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

cos x + sin x = cos 2x + sin 2x

cos x – cos 2x = sin 2x – sin x

{∵ sin A - sin B =

∴

⇒ .

∴ .

Hence,

Either,

⇒

If tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴

⇒ where m,n ϵ Z ….ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

sin x tan x – 1 = tan x – sin x

⇒ sin x tan x – tan x + sin x – 1 = 0

⇒ tan x(sin x – 1) + (sin x – 1) = 0

⇒ (sin x – 1)(tan x + 1) = 0

∴ sin x = 1 or tan x = -1

⇒ sin x = sin π/2 or tan x = tan (- π/4 )

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

and tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴ x = nπ + (-1)ⁿ (π /2) or x = mπ + (- π/4)

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

sin x tan x – 1 = tan x – sin x

⇒ sin x tan x – tan x + sin x – 1 = 0

⇒ tan x(sin x – 1) + (sin x – 1) = 0

⇒ (sin x – 1)(tan x + 1) = 0

∴ sin x = 1 or tan x = -1

⇒ sin x = sin π/2 or tan x = tan (- π/4 )

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

and tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴ x = nπ + (-1)ⁿ (π /2) or x = mπ + (- π/4)

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

3tan x + cot x = 5cosec x

⇒

⇒

⇒

⇒

⇒ {∵ sin² x + cos² x = 1}

∴

⇒ = 0

⇒

⇒

⇒

∴ cos x = -3 (neglected as cos x lies between -1 and 1)

or cos x = � (accepted value)

∴

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

3tan x + cot x = 5cosec x

⇒

⇒

⇒

⇒

⇒ {∵ sin² x + cos² x = 1}

∴

⇒ = 0

⇒

⇒

⇒

∴ cos x = -3 (neglected as cos x lies between -1 and 1)

or cos x = � (accepted value)

∴

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

∴

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0

As, cos 2x = 1 – 2sin² x and sin 2x = 2sin x cos x

∴ 3 – 2cos x – 4sin x – (1 – 2sin² x) + 2sin x cos x = 0

⇒ 2sin² x – 4sin x + 2 – 2cos x + 2sin x cos x = 0

⇒ 2(sin² x – 2sin x + 1) + 2cos x(sin x – 1) = 0

⇒ 2(sin x – 1)² + 2cos x(sin x – 1) = 0

⇒ (sin x – 1)(2cos x + 2sin x – 2) = 0

∴ sin x = 1 or sin x + cos x = 1

When, sin x = 1

We have,

sin x = sin π/2

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z

∴

When, sin x + cos x = 1

∴ { dividing by √2 both sides}

⇒ { ∵ }

⇒ { ∵ cos A cos B + sin A sin B = cos (A - B)}

If cos x = cos y, implies x = 2mπ ± y, where m ∈ Z

∴

∴

⇒

Hence,

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0

As, cos 2x = 1 – 2sin² x and sin 2x = 2sin x cos x

∴ 3 – 2cos x – 4sin x – (1 – 2sin² x) + 2sin x cos x = 0

⇒ 2sin² x – 4sin x + 2 – 2cos x + 2sin x cos x = 0

⇒ 2(sin² x – 2sin x + 1) + 2cos x(sin x – 1) = 0

⇒ 2(sin x – 1)² + 2cos x(sin x – 1) = 0

⇒ (sin x – 1)(2cos x + 2sin x – 2) = 0

∴ sin x = 1 or sin x + cos x = 1

When, sin x = 1

We have,

sin x = sin π/2

If sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z

∴

When, sin x + cos x = 1

∴ { dividing by √2 both sides}

⇒ { ∵ }

⇒ { ∵ cos A cos B + sin A sin B = cos (A - B)}

If cos x = cos y, implies x = 2mπ ± y, where m ∈ Z

∴

∴

⇒

Hence,

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

3sin² x – 5 sin x cos x + 8 cos² x = 2

⇒ 3sin² x + 3 cos² x – 5sin x cos x + 5 cos² x = 2

⇒ 3 - 5sin x cos x + 5 cos² x = 2 {∵ sin² x + cos ²x = 1 }

⇒ 5cos² x + 1 = 5sin x cos x

Squaring both sides:

⇒ (5cos² x + 1)² = (5sin x cos x)²

⇒ 25cos⁴ x + 10cos² x + 1 = 25 sin² x cos² x

⇒ 25cos⁴ x + 10cos² x + 1 = 25 (1 - cos² x) cos² x

⇒ 50cos⁴ x – 15 cos² x + 1 = 0

⇒ 50cos⁴ x – 10 cos² x – 5cos² x + 1 = 0

⇒ 10cos² x ( 5cos² x – 1) – (5cos² x – 1) = 0

⇒ (10cos² x - 1)(5cos² x – 1) = 0

∴ cos² x = 1/10 or cos² x = 1/5

Hence, when cos² x = 1/10

We have,

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

let cos α = 1/√10

∴ cos (π – α) = -1/√10

∴ x = 2nπ ± α or x = 2nπ ± (π – α)

∴ when,

When cos² x = 1/5

We have, .

If cos x = cos y, implies x = 2mπ ± y, where n ∈ Z.

let cos β = 1/√5

∴ cos (π – β) = -1/√5

∴ x = 2mπ ± β or x = 2mπ ± (π – β)

∴ when, .

…ans

Ideas required to solve the problem:

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

Given,

3sin² x – 5 sin x cos x + 8 cos² x = 2

⇒ 3sin² x + 3 cos² x – 5sin x cos x + 5 cos² x = 2

⇒ 3 - 5sin x cos x + 5 cos² x = 2 {∵ sin² x + cos ²x = 1 }

⇒ 5cos² x + 1 = 5sin x cos x

Squaring both sides:

⇒ (5cos² x + 1)² = (5sin x cos x)²

⇒ 25cos⁴ x + 10cos² x + 1 = 25 sin² x cos² x

⇒ 25cos⁴ x + 10cos² x + 1 = 25 (1 - cos² x) cos² x

⇒ 50cos⁴ x – 15 cos² x + 1 = 0

⇒ 50cos⁴ x – 10 cos² x – 5cos² x + 1 = 0

⇒ 10cos² x ( 5cos² x – 1) – (5cos² x – 1) = 0

⇒ (10cos² x - 1)(5cos² x – 1) = 0

∴ cos² x = 1/10 or cos² x = 1/5

Hence, when cos² x = 1/10

We have,

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

let cos α = 1/√10

∴ cos (π – α) = -1/√10

∴ x = 2nπ ± α or x = 2nπ ± (π – α)

∴ when,

When cos² x = 1/5

We have, .

If cos x = cos y, implies x = 2mπ ± y, where n ∈ Z.

let cos β = 1/√5

∴ cos (π – β) = -1/√5

∴ x = 2mπ ± β or x = 2mπ ± (π – β)

∴ when, .

…ans

Given,

⇒ .

On comparing both sides, we have

sin² x = cos² x = �

Note: If we want to give solution using above two equations then task will become tedious as sin x can be positive at that time cos will be negative and similar 4-5 cases will arise. So inspite of combining all solutions at the end,we proceed as follows

combining both we can say that,

all the solutions of first 2 equations combined will satisy this single equation

tan² x = 1

tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴

Given,

⇒ .

On comparing both sides, we have

sin² x = cos² x = �

Note: If we want to give solution using above two equations then task will become tedious as sin x can be positive at that time cos will be negative and similar 4-5 cases will arise. So inspite of combining all solutions at the end,we proceed as follows

combining both we can say that,

all the solutions of first 2 equations combined will satisy this single equation

tan² x = 1

tan x = tan y, implies x = nπ + y, where n ∈ Z.

∴

sin x + 1 = 2 × (cos x)²

sin x + 1 = 2 × (1 - (sin x)²)

sin x + 1 = 2 – 2(sin x)²

2(sin x)² + sin x - 1 = 0

Consider a=sin x

So, the equation will be

2a²+a-1=0

From the equation a=0.5 or -1

Which implies

Sin x=0.5 or sin x=(-1)

Therefore x=30° or 270°

But for x=270° our equation will not be defined as cos (270° )=0

So, the solution for x=30°

According to trigonometric equations

If sin x=sin a

Then x=nπ – na

Here sin x=sin30

So, x=nπ + (-1)ⁿ × 30

For n=0, x=30 and n=1,x=150° and for n=2,x=390

Hence between 0 to 2π there are only 2 possible solutions.

4sin x – 3cos x = 7

4sin x – 7 = 3cos x

Squaring both sides

16(sin x)² + 49 – 56sin x = 9(cos x)²

16(sin x)² + 49 – 56sin x = 9((sin x)²-1)

16(sin x)²-9(sin x)²-56sin x+49+9=0

7(sin x)²-56sin x+58=0

Solving the quadratic equation

Sin x = 6.7774 or 1.2225

But we know that sinθ lies between [-1,1]

So there are no solutions for this given equation

sin²2x = cos²2x

sin ²2x = 1- Sin²2x

2 sin ²2x = 1

sin 2x=sin45

So

cos30°sin x – sin30°cos x =a

sin (x-30)=a

As the range of sin function is from [-1,1]

So the value of a can be R-[-1,1]

i.e. a ∈ (-∞, -2) ∪ (2, ∞)

As cos x = cos θ

Then x=2nπ ± θ

And it is said that it has exactly one solution.

So θ=0 and

=nπ

In the given interval taking n=1,x=π {n=0 is not possible as cos 0 = 1 not -1 but cos π is -1}

2y=1

i.e.

and y = cos x

so, to get the intersection points we must equate both the equations

i.e.

so, cos x = cos 60°

and we know if cos x = cos a

then x=2nπ ± a where a ϵ [0, π]

so here

So the possible values which belong [0,2π] are .

There are a total of 2 points of intersection.

a, a+r, a+2r

so A₁+A₃=2A₂

here sin2x + cos2x = 1

2sin x cos x + 1-2sin²x =1

sin x cos x - sin²x=0

sin x (cos x-sin x)=0

if sin x =0

then x =0, π

if sin x=cos x

then x = π/4

So, all possible values are

Y=cosec x and

So

Sin x = -2

Which is not possible

So

There are 0 points of intersection.

8 cos²x+10 cos x+4 cos x+5=0

8 cos²x+14 cos x+5=0

Solving the quadratic equation, we get,

cos x = -0.5

cos x = cos 120°

So

And when n=1

1-cos²x-cos x=0.25

cos x ²x+cos x-0.75=0

Solving the quadratic equation we get

cos x = 0.5

cos x = cos60°

x=60° when n=0

And x=300° when n=1

Let tan (15°) = tan(45°-30°)

We know that

We now

And

So, 3 tan (x - 15^o) = tan (x + 15^o) can be written as follows

(3 tan x – 3tan15)(1-tan x × tan15) = (1+tan x × tan15)(tan x + tan15)

3 tan x – 3 tan15-3 tan²x tan(15-3) tan x tan²15 = tan x + tan15 + tan²x tan15 + tan x tan²15

Solving the equation,

And putting

We get tan x - 1 = 0

Therefore, tan x =1

So, x=45°

Or

2sin²x=3cos x

2-2cos²x=3cos x

Solving the quadratic equation, we get

cos x= 1/2

Therefore x=60° and 300°

i.e.

cos 5x = -cos x

cos 5x+cos x = 0

We know

Here

Now from the above equation it would be,

2cos 3x cos 2x=0

cos 3x cos 2x=0

cos 3x=0 or cos 2x=0

for cos3x=0

for cos2x=0

so the values of the x less than equal to 90° are

√3+√3 tan²x = 4 tan x

√3 tan²x-4 tan x+√3=0

Therefore

Therefore

But here the smallest angle is π /6

Option C

cos ²x = (2-√3 sin x)²

1-sin²x = 4+3 sin²x-4√3 sin x

4 sin²x-4√3 sin x+3=0

Option A

Tan x=tana

X=nπ+a

So tan px – tan qx = 0

tan px = tan qx

px = nπ + qx

(p-q)x=nπ

Here in this series

So, this is in AP.

Option A

1-tan²x=a tan x

tan²x+a tan x – 1 = 0

As it is given a be any real number take a=0,

For a=0

Tan x = +1 or -1

In first quadrant only tan(π/4)=1

So, there is only one root that lies in the first quadrant.

Option C