FIND THE EXACT SOLUTIONS OF THE GIVEN TRIGONOMETRIC EQUATION

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

Practice Questions

Question 1 :

Solve the following equation :

cos θ + cos 3θ  =  2cos 2θ

Solution :

cos θ + cos3θ = 2cos2θ

Let us use the formula for cos C + cos D

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

cos θ + cos3θ = 2cos2θ

2 cos 2θ cos θ - 2 cos 2θ = 0

2 cos 2θ (cos θ - 1)  =  0

2 cos 2θ  =  0         cos θ - 1  =  0 

2 cos 2θ  =  0 

 cos 2θ  =  0 

2θ  =  cos-1(0)

2θ  =  (2n + 1)π/2

θ  =  (2n + 1)π/4

cos θ - 1  =  0 

cos θ  =  1

α = 0

θ  =  2nπ ± α, n ∈ Z

θ  =  2nπ ± 0, n ∈ Z

θ  =  2nπ , n ∈ Z

So, the solution is {(2n + 1)π/4, 2nπ}.

Question 2 :

Solve the following equation :

sin θ + sin 3θ + sin 5θ  =  0

Solution :

sin θ + sin 3θ + sin 5θ  =  0

First let us use the formula for sin C + sin D

2 sin 3θ cos 2θ + sin 3θ = 0

sin 3θ (2cos 2θ + 1)  =  0

sin 3θ  =  0  (or)  2cos 2θ + 1  =  0

sin 3θ  =  0     

3θ = nπ; n ∈ Z

θ = nπ/3; n ∈ Z

2cos 2θ + 1  =  0

2cos 2θ  =  -1

cos 2θ  =  -1/2

a = π - (π/3)  ==> 2π/3

2θ  =  2nπ ± α, n ∈ Z

2θ  =  2nπ ± 2π/3, n ∈ Z

θ  =  nπ ± π/3, n ∈ Z

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