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TRIGONOMETRIC EQUATION Ex.16 If sin 5x + sin 3x + sinx = 0 and 0 ≤ x ≤ π/2, then x is equal to [1] π/12 [2] π/6 [3] π/4 Sol. sin 5x + sinx = – sin 3x ⇒ 2 sin 3x cos 2x + sin 3x = 0 ⇒ sin 3x (2 cos 2x + 1) = 0 ⇒ sin 3x = 0, cos 2x = – 1/2 ⇒ x = nπ, x = nπ ± (π/3) So x = π/3 [4] π/3 Ans. [4] Ex.17 The number of solutions of equation, sin 5x cos 3x = sin 6x cos 2x, in the interval [0, π] are [1] 3 [2] 4 [3] 5 [4] 6 Sol. The given equation can be written as 1 1 (sin 8x + sin 2x) = (sin 8x + sin 4x) 2 2 or, sin 2x – sin 4x ⇒ – 2 sin xcos 3x = 0 Hence sin x = 0 or cos 3x = 0. That is, x = nπ(n ∈ Ι), or 3x = kπ + x ∈ [0, π], then given equation is satisfied if x = 0, π, π (k ∈ Ι). Therefore, since 2 π π 5π , or 6 2 6 Ans. [3] Ex.18 Solve the following system of equation sinx + cosy =1 , cos2x – cos2y = 1 Sol. Given sinx + cosy = 1 ......(i) and (1 – 2sin2x) – (2cos2y –1) = 1 ⇒ sin2x + cos2y = 1/2 Put sin x = u and cosy = v in (i) and (ii) u + v = 1 ⇒ u2 + v2 = 1/ 2 solving above equations u = π 1 1 1 and v = , n ∈I ⇒ sin x = ⇒ x = nπ + (–1)n 6 2 2 2 π 1 ⇒ y = 2mπ ± , m ∈ I ⇒ ∴ the given equation have solution 3 2 π π x = nπ + (–1)n , n ∈ I and y = 2mπ ± , m ∈ I 6 3 cosy = .....(ii)