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The solution set of $\sin 7 x + \sin 3 x = \sin 2 x + \sin 8 x$ is $(n \in z)$

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{(2 n + 1) \frac{π}{3}}

\frac{n π}{5}

{(2 n + 1) \frac{π}{9}}

{(2 n + 1) \frac{π}{7}}

answer is A.

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Detailed Solution

\sin 7 x + \sin 3 x = \sin 2 x + \sin 8 x

$2 \sin 5 x . \cos 2 x = 2 \sin 5 x . \cos 3 x$

$2 \sin 5 x . \cos 2 x - 2 \sin 5 x . \cos 3 x = 0$

$2 \sin 5 x (\cos 2 x - \cos 3 x) = 0$

$- 4 \sin 5 x \sin \frac{5 x}{2} \sin \frac{x}{2} = 0$

$\sin 5 x = 0$ (or) $\sin \frac{5 x}{2} = 0$ (or) $\sin \frac{x}{2} = 0$

$5 x = n π$ (or) $\frac{5 x}{2} = n π$ (or) $\frac{x}{2} = n π$

$x = \frac{n π}{5}$ (or) $x = \frac{2 n π}{5}$ (or) $x = 2 n π$

Common solution $x = \frac{n π}{5}$

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