The sum of all values of $x$ in $[0,2\pi ]$, for which $\sin x+\sin 2x+\sin 3x+\sin 4x=0$, is equal to :

Given,

$\sin x+\sin 2x+\sin 3x+\sin 4x=0$

$\Rightarrow (\sin x+\sin 4x)+(\sin 2x+\sin 3x)=0$

$\Rightarrow 2\sin \frac{5x}{2}\cos \frac{3x}{2}+2\sin \frac{5x}{2}\cos \frac{x}{2}=0$

$\Rightarrow 2\sin \frac{5x}{2}\left\{\cos \frac{3x}{2}+\cos \frac{x}{2}\right\}=0$

$\Rightarrow 2\sin \frac{5x}{2}\left\{2\cos x\cos \frac{x}{2}\right\}=0$

$2\sin \frac{5x}{2}=0\Rightarrow \frac{5x}{2}=0,\pi ,2\pi ,3\pi ,4\pi ,5\pi$

$\Rightarrow x=0,\frac{2\pi }{5},\frac{4\pi }{5},\frac{6\pi }{5},\frac{8\pi }{5},2\pi$

and $\cos \frac{x}{2}=0\Rightarrow \frac{x}{2}=\frac{\pi }{2}\Rightarrow x=\pi$

and $\cos x=0\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2}$

So sum $=6\pi +\pi +2\pi =9\pi$