Find the general solution of the equation $\sin x + \sin 3x + \sin 5x = 0$.

Given equation: $\sin x + \sin 3x + \sin 5x = 0$
Rearranging the terms: $(\sin 5x + \sin x) + \sin 3x = 0$
Using the formula $\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$:
$2 \sin \left(\frac{5x+x}{2}\right) \cos \left(\frac{5x-x}{2}\right) + \sin 3x = 0$
$2 \sin 3x \cos 2x + \sin 3x = 0$
Taking $\sin 3x$ as a common factor:
$\sin 3x (2 \cos 2x + 1) = 0$
This gives two cases:
Case $1$: $\sin 3x = 0$
$3x = n\pi$,where $n \in \mathbb{Z}$
$x = \frac{n\pi}{3}$,where $n \in \mathbb{Z}$
Case $2$: $2 \cos 2x + 1 = 0$
$\cos 2x = -\frac{1}{2}$
Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have $\cos 2x = -\cos \frac{\pi}{3} = \cos \left(\pi - \frac{\pi}{3}\right) = \cos \frac{2\pi}{3}$
$2x = 2n\pi \pm \frac{2\pi}{3}$,where $n \in \mathbb{Z}$
$x = n\pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$
Thus,the general solution is $x = \frac{n\pi}{3}$ or $x = n\pi \pm \frac{\pi}{3}$,where $n \in \mathbb{Z}$.