If $sin x + sin 3 x + sin 5 x = 0$ , then the solution is

x = \frac{2 n π}{3}, n ϵ

I

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\frac{n π}{3}, n ϵ

I

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(3 n \pm 1) \frac{π}{3} n ϵ

I

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none of these

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Solution

The correct option is D none of these
$sin x + sin 3 x + sin 5 x = 0$

$\Rightarrow (sin 5 x + sin x) + sin 3 x = 0$

$\Rightarrow 2 sin 3 x cos 2 x + sin 3 x = 0$

$\Rightarrow sin 3 x (2 cos 2 x + 1) = 0$

$sin 3 x = 0$ or $2 cos 2 x + 1 = 0$

$sin 3 x = 0$ or, $cos 2 x = - \frac{1}{2}$

Now, $sin 3 x = 0 ⟹ 3 x = n π ⟹ x = \frac{n π}{3}, n \in Z$

And, $cos 2 x = - \frac{1}{2}$

$cos 2 x = cos \frac{2 π}{3}$

$2 x = 2 m π \pm \frac{2 π}{3}, m \in Z$

$x = m π \pm \frac{π}{3}, m \in Z$

Hence, the general solution of the given equation is :
$x = \frac{n π}{3}$ or, $x = m π \pm \frac{π}{3}$ , where $m, n \in Z$

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