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

**Question 1 :**

Solve the following equations:

(iii) cos θ + cos3θ = 2cos2θ

**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 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 |

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

(iv) sin θ + sin3θ + sin5θ = 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 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 |

After having gone through the stuff given above, we hope that the students would have understood how to find the exact solution of a trigonometric equation.

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