$\sin x+\sin 3 x+\sin 5 x=0$

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

$\Rightarrow\left[2 \sin \left(\frac{x+5 x}{2}\right) \cos \left(\frac{x-5 x}{2}\right)\right]+\sin 3 x=0$ $\left[\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

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

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

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

$\Rightarrow \sin 3 x=0 \quad$ or $\quad 2 \cos 2 x+1=0$

Now, $\sin 3 x=0 \Rightarrow 3 x=n \pi,$ where $n \in Z$

i.e., $x=\frac{n \pi}{3},$ where $n \in Z$

$2 \cos 2 x+1=0$

$\Rightarrow \cos 2 x=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)$

$\Rightarrow \cos 2 x=\cos \frac{2 \pi}{3}$

$\Rightarrow 2 x=2 n \pi \pm \frac{2 \pi}{3},$ where $n \in Z$

$\Rightarrow x=n \pi \pm \frac{\pi}{3},$ where $n \in Z$

Therefore, the general solution is $\frac{n \pi}{3}$ or $n \pi \pm \frac{\pi}{3}, n \in Z$