**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) θ = 2nπ ± α, n ∈ Z θ = nπ + α, n ∈ Z |

**Question 1 :**

Solve the following equations:

(i) sin 5x − sin x = cos3x

**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 3x = (2n + 1)π/2 x = (2n + 1)(π/6) |
2 sin 2x - 1 = 0 sin 2x = 1/2 2x = sin where a = π/6 α ∈ [−π/2, π/2] θ = nπ + (−1) 2x = nπ + (−1) x = (nπ/2) + (−1) |

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

(ii) 2 cos^{2} θ + 3sinθ − 3 = 0

**Solution :**

2 cos^{2} θ + 3sinθ − 3 = 0

2(1 - sin^{2} θ) + 3sinθ − 3 = 0

2 - 2sin^{2} θ + 3sinθ − 3 = 0

-2sin^{2} θ + 3sinθ − 1 = 0

2sin^{2} θ - 3sinθ + 1 = 0

Let t = sinθ

2t^{2} - 3t + 1 = 0

(t - 1) (2t - 1) = 0

t - 1 = 0 t = 1 sin θ = 1 θ = sin a = π/2 θ = nπ + (−1) θ = nπ + (−1) |
2t - 1 = 0 2t = 1 t = 1/2 sin θ = 1/2 a = π/6 θ = nπ + (−1) |

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

After having gone through the stuff given above, we hope that the students would have understood how to solve trigonometric equations.

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