FIND ALL SOLUTIONS OF THE EQUATION EXPRESS THE SOLUTIONS IN RADIANS

The equations containing trigonometric functions of unknown angles are known as trigonometric equations. A solution of trigonometric equation is the value of unknown angle that satisfies the equation.

General Solution :

The solution of a trigonometric equation giving all the admissible values obtained with the help of periodicity of a trigonometric function is called the general solution of the equation.

Trigonometric equation

sin θ = 0

cos θ = 0

tan θ = 0

sin θ = sinα, where

α ∈ [−π/2, π/2]

cos θ = cos α, where α ∈ [0,π]

tan θ = tanα, where

α ∈ (−π/2, π/2)

General solution

θ = nπ; n ∈ Z

θ = (2n + 1) π/2; n ∈ Z

θ = nπ; n ∈ Z


θ = nπ + (−1)n α, n ∈ Z

θ = 2nπ ± α, n ∈ Z


θ = nπ + α, n ∈ Z

Question 1 :

Solve the following equations:

(v) sin 2θ − cos 2θ − sin θ + cos θ = 0

Solution :

sin 2θ − cos 2θ − sin θ + cos θ  =  0

sin 2θ − sin θ + cos θ − cos 2θ  =  0

Let us use the formula for sin C - sin D and cos C - cos D

 sin C - sin D  =  2 cos (C + D)/2 sin (C - D)/2

cos C - cos D  =  2 sin (C + D)/2 sin (C - D)/2

sin 2θ − sin θ  =  2 cos 3θ/2 sin θ/2  -----(1)

cos θ − cos 2θ  =  2 sin 3θ/2 sin θ/2  -----(2)

(1) + (2)

   =  2 cos 3θ/2 sin θ/2 + 2 sin 3θ/2 sin θ/2

  =  2 sin θ/2 [cos 3θ/2 + sin 3θ/2]  

sin θ/2  =  0 

 sin θ/2  =  0

θ/2  =  nπ

θ  =  2nπ

cos 3θ/2 + sin 3θ/2  =  0

cos 3θ/2  =  - sin 3θ/2 

sin 3θ/2/cos 3θ/2  =  -1

tan 3θ/2  =  -1

a = - π/4

 θ  =  nπ + a

3θ/2  =  nπ - π/4

3θ  =  2nπ - π/2

θ  =  2nπ/3 - π/6

Hence the solution is {2nπ, 2nπ/3 - π/6}.

(vi)  sin θ + cos θ = 2

Solution :

sin θ + cos θ = 2

Divide by 2 on both sides

(1/2) sin θ + (1/2) cos θ = 1

cos (π/4) cos θ - sin (π/4) sin θ = cos 0

cos (π/4 - θ)  =  cos 0

θ = 2nπ + a

θ = 2nπ + π/4

θ = (8n + 1)π/4

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