General solution of the trigonometric equation
$sin x + sin 5 x = sin 2 x + sin 4 x$ is

\frac{2 n π}{3}

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\frac{n π}{3}

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2 n π

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n π

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Solution

The correct option is B $\frac{n π}{3}$
Given: $sin x + sin 5 x = sin 2 x + sin 4 x$

Using $sin C + sin D = 2 sin (\frac{C + D}{2}) cos (\frac{C - D}{2})$
$\Rightarrow 2 sin 3 x . cos 2 x = 2 sin 3 x . cos x$
$\Rightarrow sin 3 x . cos 2 x - sin 3 x . cos x = 0$
$\Rightarrow sin 3 x . (cos 2 x - cos x) = 0$
Either $sin 3 x = 0 or cos 2 x = cos x$

Case:1
$sin 3 x = 0 \Rightarrow 3 x = n π \Rightarrow x = \frac{n π}{3} \forall n \in Z$

Case: 2
$cos 2 x = cos x 2 x = 2 n π \pm x$
$x = 2 n π, \frac{2 n π}{3} \forall n \in Z$

From Case:1 and Case: 2, combined solution is
$x = \frac{n π}{3} \forall n \in Z$

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