Trigonometric Equations

To Find: Principal solution.

[NOTE: The solutions of a trigonometry equation for which 0x2 is called principal solution]

(i) Given:

Formula used: sin = sin = n + (-1)ⁿ , nI

By using above formula, we have

= sin x = n +(-1)ⁿ

Put n= 0 x = 0 +(-1)⁰ x =

Put n= 1 x = 1 +(-1)¹ x = ¹ x = =

So principal solution is x= and

(ii) Given:

Formula used: cos = cos = 2n , nI

By using above formula, we have

= cos = 2n, nI

Put n= 0 x = 2n x =

Put n= 1 x = 2 x = , x = ,

[ 2 So it is not include in principal solution]

So principal solution is x= and

(iii) Given:

Formula used: tan = tan = n , nI

By using above formula, we have

= tan x = n, nI

Put n= 0 x = n x =

Put n= 1 x = x = x =

So principal solution is x= and

(iv) Given:

We know that tan cot = 1

So cotx = tanx =

The formula used: tan = tan = n , nI

By using the above formula, we have

tanx = = tan = n, nI

Put n= 0 x = n x =

Put n= 1 x = x =

So principal solution is x= and

(v) Given: cosec x = 2

We know that cosec sin = 1

So sinx =

Formula used: sin = sin = n + (-1)ⁿ , n

By using above formula, we have

sinx = = sin = n +(-1)ⁿ

Put n= 0 = 0 +(-1)⁰ =

Put n= 1 = 1 +(-1)¹ = ¹ = =

So principal solution is x= and

(vi) Given: sec x =

We know that sec cos = 1

So cosx =

Formula used: cos = cos = 2n , nI

By using the above formula, we have

cosx = = cos x = 2n, nI

Put n= 0 x = 2n x =

Put n= 1 x = 2 x = , x = ,

[ 2 So it is not include in principal solution]

So principal solution is x= and

To Find: Principal solution.

(i) Given:

Formula used: sin = sin = n + (-1)ⁿ , nI

By using above formula, we have

= -sin = sin(= sin x = n +(-1)ⁿ

Put n= 0 x = 0 +(-1)⁰ x =

Put n= 1 x = 1 +(-1)¹ x = ¹ x = =

[ NOTE: = ]

So principal solution is x= and

(ii) Given: cosx =

Formula used: cos = cos = 2n , nI

By using above formula, we have

cosx = = cos x = 2n, nI

Put n= 0 x = 2 × 0 × x =

Put n= 1 x = 2 x = , x = ,

[ 2 So it is not include in principal solution]

So principal solution is x= and

(iii) Given: tan x = -1

Formula used: tan = tan = n , nI

By using above formula, we have

tan x = -1 = tan x = n, nI

Put n= 0 x = n x =

Put n= 1 x = x = x =

So principal solution is x= and

(iv) Given: cosec x =

We know that cosec sin = 1

So sinx =

Formula used: sin = sin = n + (-1)ⁿ , n

By using above formula, we have

sinx = = sin = n +(-1)ⁿ

Put n= 0 x = 0 +(-1)⁰ x =

Put n= 1 x = 1 +(-1)¹ x = ¹ x = =

[ NOTE: = ]

So principal solution is x= and

(v) Given: tan x = -

Formula used: tan = tan = n , nI

By using above formula, we have

tan x = - = tan x = n, nI

Put n= 0 x = n x =

Put n= 1 x = x =

So principal solution is x= and

(vi) Given: sec x =

We know that sec cos = 1

So cosx =

Formula used: cos = cos = 2n , nI

By using the above formula, we have

cosx = = cos x = 2n, nI

Put n= 0 x = 2n x =

Put n= 1 x = 2 x = , x = ,

[ 2 So it is not include in principal solution]

So principal solution is x= and

To Find: General solution.

[NOTE: A solution of a trigonometry equation generalized by means of periodicity, is known as general solution]

(i) Given: sin 3x = 0

Formula used: sin= 0 = n , nI

By using above formula, we have

sin 3x = 0 3x = n x = where nI

So general solution is x= where nI

(ii) Given: sin = 0

Formula used: sin= 0 = n , nI

By using above formula, we have

sin = 0 = n x = where nI

So general solution is x= where nI

(iii) Given: sin = 0

Formula used: sin= 0 = n , nI

By using the above formula, we have

sin = 0 = n x = n- where nI

So general solution is x= n- where nI

(iv) Given: cos 2x = 0

Formula used: cos= 0 = (2n+1) , nI

By using above formula, we have

cos 2x = 0 2x = (2n+1) x = (2n+1) where nI

So general solution is x= (2n+1)where nI

(v) Given: cos = 0

Formula used: cos= 0 = (2n+1) , nI

By using the above formula, we have

cos = 0 = (2n+1) x = (2n+1) where nI

So general solution is x= (2n+1)where nI

(vi) Given: cos = 0

Formula used: cos= 0 = (2n+1) , nI

By using the above formula, we have

cos = 0 = (2n+1) x = (2n+1) - x = n + where nI

So general solution is x= n + where nI

(vii) Given: tan 2x = 0

Formula used: tan= 0 = n , nI

By using above formula, we have

tan 2x = 0 2x = n x = where nI

So general solution is x= where nI

(viii) Given: tan = 0

Formula used: tan= 0 = n , nI

By using above formula, we have

tan = 0 = n 3x = n - x = - where nI

So general solution is x = - where nI

(ix) Given: tan = 0

Formula used: tan= 0 = n , nI

By using above formula, we have

tan = 0 = n 2x = n - x = + where nI

So general solution is x = + where nI

To Find: General solution.

(i) Given: sin x =

Formula used: sin = sin = n + (-1)ⁿ , nI

By using above formula, we have

sin x = = sin x = n + (-1)ⁿ .

So general solution is x = n + (-1)ⁿ . where nI

(ii) Given: cos x = 1

Formula used: cos = cos = 2n , nI

By using above formula, we have

cos x = 1= cos(0) x = 2n , nI

So general solution is x = 2n where nI

(iii) Given: sec x =

We know that sec cos = 1

So cosx =

Formula used: cos = cos = 2n , nI

By using above formula, we have

cosx = = cos x = 2n , nI

So general solution is x = 2n where nI

To Find: General solution.

(i) Given: cos x =

Formula used: cos = cos = 2n , nI

By using above formula, we have

cos x = = cos()= cos()=cos() x = 2n , nI

So general solution is x = 2n where nI

(ii) Given: cosec x =

We know that cosec sin = 1

So sinx =

Formula used: sin = sin = n + (-1)ⁿ , n

By using above formula, we have

sinx = = sin x = n +(-1)ⁿ.

So general solution is x = n +(-1)ⁿ. where nI

(iii) Given: tan x = -1

Formula used: tan = tan = n , nI

By using above formula, we have

tan x = -1= tan x = n, nI

So the general solution is x = nwhere nI

To Find: General solution.

(i) Given: sin 2x =

Formula used: sin = sin = n + (-1)ⁿ , nI

By using above formula, we have

sin 2x = = sin 2x = n + (-1)ⁿ . x = + (-1)ⁿ . , nI

So general solution is x = + (-1)ⁿ . where nI

(ii) Given: cos 3x =

Formula used: cos = cos = 2n , nI

By using above formula, we have

cos 3x = = cos() 3x = 2n x = , nI

So the general solution is x = where nI

(iii) Given: tan =

Formula used: tan = tan = n , nI

By using above formula, we have

tan = = tan = n x = , nI

So general solution is x =(3n+1),where nI

To Find: General solution.

(i) Given: sec 3x = -2

We know that sec cos = 1

So cos 3x =

Formula used: cos = cos = 2n , nI

By using above formula, we have

cos 3x = = -cos= cos= cos 3x = 2n x = , nI

So the general solution is x = , ,where nI

(ii) Given: cot 4x = -1

We know that tan cot = 1

So tan 4x = -1

Formula used: tan = tan = n , nI

By using above formula, we have

tan 4x = -1= tan 4x = n x = , nI

So general solution is x = (4n+3) ,where nI

(iii) Given: cosec 3x =

We know that cosec sin = 1

So sin 3x =

Formula used: sin = sin = n + (-1)ⁿ. , nI

By using above formula, we have

sin 3x = = sin 3x= n +(-1)ⁿ . x= +(-1)ⁿ . , nI

So general solution is x = +(-1)ⁿ . , where nI

To Find: General solution.

(i) Given: 4cos² x = 1 cos² x =

cos² x = cos²

Formula used: cos² = cos² = n , nI

By using the above formula, we have

x = n , nI

So the general solution is x = n where nI

(ii) Given: 4sin² x – 3 = 0 sin² x = = sin²

sin² x = sin²

Formula used: sin² = sin² = n , nI

By using the above formula, we have

x = n , nI

So the general solution is x = n where nI

(ii) Given: tan² x = 1 tan² x = tan²

tan² x = tan²

The formula used: tan² = tan² = n , nI

By using the above formula, we have

x = n , nI

So the general solution is x = n where nI

To Find: General solution.

(i) Given: cos 3x = cos 2x cos 3x - cos 2x = 0 -2sin sin = 0

[NOTE: cos C – cos D = -2sin sin ]

So, sin = 0 or sin= 0

Formula used: sin = 0 = n , nI

= n or = m where n, m I

x = 2 n/5 or x = 2m where n, m I

So general solution is x = 2 n/5 or x = 2m where n, m I

(ii) Given: cos 5x = sin 3x cos 5x = cos

Formula used: cos = cos = 2n , nI

By using the above formula, we have

5x = 2n or 5x = 2n

8x = 2n or 2x = 2n

x = or x = n where n I

So general solution is x = or x = n where n I

(iii) Given: cos mx = sin nx cos mx = cos

Formula used: cos = cos = 2k , kI

By using the above formula, we have

mx = 2k or 5x = 2k

(m+n)x = 2k or (m-n)x = 2k

x = or x = where k I

x = or x = where k I

So the general solution is x = or x = where k I

To Find: General solution.

Given: sin x = tan x sin x = sin x cos x

So sin x = 0 or cos x = 1 = cos(0)

Formula used: sin= 0 = n, nI and cos = cos = 2k , kI

x = n or x = 2k where n, k I

So general solution is x =n or x = 2k where n, k I

To Find: General solution.

Given: 4sin x cos x + 2sin x + 2cos x + 1 = 0 2sin x(2cos x + 1) + 2cos x + 1 = 0

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

cos x = = cos() or sin x = = sin

Formula used: cos = cos = 2n or sin = sin = m + (-1)^m where n,mI

x = 2n or x = m +(-1)^m . where n, mI

So the general solution is x =2n or x = m +(-1)^m . where n, mI

To Find: General solution.

Given: sec² 2x = 1- tan 2x 1 + tan²2x+ tan 2x = 1 tan 2x (1+tan 2x) = 0

So, tan 2x = 0 or tan 2x = -1 = tan ()

Formula used: : tan= 0 = n , nI and tan = tan = k , kI

By using above formula, we have

2x = n or 2x = k x = or x =

So the general solution is x = or x = where n, kI

To Find: General solution.

Given: tan³ x – 3tan x = 0 tan x(tan² x – 3) = 0 tan x = 0 or tanx =

tan x = 0 or tanx = tan() or tan x = tan()

Formula used: tan= 0 = n , nI, tan = tan = k , kI

So x = n or x = k + or x = p + where n, k, pI

So general solution is x = n or x = k + or x = p + where n, k, pI

To Find: General solution.

Given: sin x + sin 3x + sin 5x = 0 sin 3x + 2sin 3x cos 2x= 0 sin 3x (1 + 2cos 2x) = 0

[NOTE: sin C + sin D = 2sin (C+D)/2 × cos (C-D)/2]

sin 3x = 0 or cos 2x = = cos()

Formula used: sin= 0 = n , nI, cos = cos = 2k , kI

3x = n or 2x = 2k x = or x = k where n,k I

So general solution is x = or x = k where n, k, I

To Find: General solution.

Given: sin x tan x – 1 = tan x – sin x sin x(tan x + 1) = tan x + 1

So sin x = 1 = sin () or tan x = -1 = tan()

Formula used: sin = sin = n + (-1)ⁿ , nI and tan = tan = k , kI

x = n + (-1)ⁿ or x = k where n, k I

So general solution is x = n + (-1)ⁿ or x = k where n, k, I

To Find: General solution.

Given: cos x + sin x = 1 cos(x - ) = = cos

[divide on both sides and cos(x-y) = cos x cos y - sin x sin y]

Formula used: cos = cos = 2k , kI

x - = 2k x = 2k + x = 2k + or x = 2k +

x = 2k + or x = 2k

So general solution is x = 2n + or x = 2n where n I

To Find: General solution.

Given: cos x - sin x = 1 cos(x + ) = = cos

[divide on both sides and cos(x-y) = cos x cos y - sin x sin y]

So sin x = 0 or cos x = 0

Formula used: cos = cos = 2k , kI

x + = 2k x = 2k - x = 2k - or x = 2k -

x = 2k - or x = 2k

So general solution is x = 2n + or x = (2n-1) where n I

To Find: General solution.

Given: cos x + sin x = 1 cos (x - ) = = cos( or cos()

[Divide on both sides and cos(x-y) = cos x cos y - sin x sin y]

Formula used: cos = cos = 2n

By using above formula, we have

x - = 2n x = 2n +

x = 2n + or x = 2n - where n I

So general solution is x = 2n + or x = 2n - where n I

To Find: General solution.

Given: 2 tan x – cot x + 1 = 0 2tan²x – 1 + tan x = 0 2tan²x – 1 + 2tan x – tanx = 0 2tanx(tanx +1) – (1+ tanx) = 0

(2tanx-1) (1+ tanx) = 0 tan x = = or tan x = -1 = tan

Formula used: tan = tan = n , nI

x = n or x = n

So the general solution is x = nor x = n where n I

To Find: General solution.

Given: sin x tan x – 1 = tan x – sin x sin x(tan x + 1) = tan x + 1

So sin x = 1 = sin () or tan x = -1 = tan()

Formula used: sin = sin = n + (-1)ⁿ , nI and tan = tan = k , kI

x = n + (-1)ⁿ or x = k where n, k I

So general solution is x = n + (-1)ⁿ or x = k where n, k I

To Find: General solution.

Given: cot x + tan x = 2 cosec x cos²x + sin²x = 2 sinx cosx cosec x 1 = sin 2x cosec x

cosec 2x = cosecx sin x = sin 2x sin x = 2 sin x cos x sin x = 0 or cos x = = cos()

Formula used: sin = 0 = n, cos = cos = 2n

By using above formula , we have

x = n or x = 2m where n, mI

So general solution is x = n or x = 2m where n, mI