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 equationsin θ = 0cos θ = 0tan θ = 0sin 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 solutionθ = nπ; n ∈ Zθ = (2n + 1) π/2; n ∈ Zθ = nπ; n ∈ Zθ = nπ + (−1)n α, n ∈ Zθ = 2nπ ± α, n ∈ Zθ = 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/2θ  =  (2n + 1)π/4 cos θ - 1  =  0 cos θ  =  1α = 0θ  =  2nπ ± α, n ∈ Zθ  =  2nπ ± 0, n ∈ Zθ  =  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 ∈ Zθ = nπ/3; n ∈ Z 2cos 2θ + 1  =  02cos 2θ  =  -1cos 2θ  =  -1/2a = π - (π/3)  ==> 2π/32θ  =  2nπ ± α, n ∈ Z2θ  =  2nπ ± 2π/3, n ∈ Zθ  =  nπ ± π/3, n ∈ Z

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