SOLVING TRIGONOMETRIC EQUATIONS EXAMPLES

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

Example 1 :

Solve the following equation :

sin 5x - sin x  =  cos 3x  

Solution :

sin 5x − sin x  =  cos3x

Let us use the formula for sin C - sin D.

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

2 cos 3x sin 2x  =  cos 3x

2 cos 3x sin 2x - cos 3x  =  0

cos 3x (2 sin 2x  - 1)  =  0

cos 3x   =  0

3x  =  cos-1(0)

3x  =  (2n + 1)π/2

x  =  (2n + 1)(π/6)

2 sin 2x  - 1  =  0

sin 2x  =  1/2

2x = sin-1 (1/2) 

where a = π/6 α ∈ [−π/2, π/2]

θ = nπ + (−1)n α, n ∈ Z

2x = nπ + (−1)n π/6, n ∈ Z

x = (nπ/2) + (−1)n π/12, n ∈ Z

So, the solution is { (2n + 1)(π/6), (nπ/2) + (−1)n π/12}.

Example 2 :

Solve the following equation :

2 cos2 θ + 3sinθ - 3  =  0

Solution :

2 cos2 θ + 3sinθ - 3  =  0

2(1 - sin2 θ) + 3sinθ - 3 = 0

2 - 2sin2 θ + 3sinθ − 3 = 0

-2sin2 θ + 3sinθ − 1 = 0

2sin2 θ - 3sinθ + 1 = 0

Let t = sinθ

2t2 - 3t + 1  =  0

(t - 1)(2t - 1)  =  0

t - 1  =  0

t  =  1

 sin θ  =  1

θ  =  sin-1 (1)

a  =  π/2

θ  =  nπ + (−1)n α, n ∈ Z

θ  = nπ + (−1)n (π/2), n ∈ Z

2t - 1  =  0

2t  =  1

t  =  1/2

 sin θ  =  1/2

a  =  π/6

θ = nπ + (−1)n (π/6), n ∈ Z

So, the solution is {nπ + (−1)n (π/2) , nπ + (−1)n (π/6)}.

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