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.
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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
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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
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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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