The sum of all values of $x$ in $[0,2 \pi]$, for which $\sin x+\sin 2 x+\sin 3 x+\sin 4 x=0$, is equal to :
Correct Option: , 4
$(\sin x+\sin 4 x)+(\sin 2 x+\sin 3 x)=0$
$\Rightarrow 2 \sin \frac{5 x}{2}\left\{\cos \frac{3 x}{2}+\cos \frac{x}{2}\right\}=0$
$\Rightarrow 2 \sin \frac{5 x}{2}\left\{2 \cos x \cos \frac{x}{2}\right\}=0$
$2 \sin \frac{5 x}{2}=0 \Rightarrow \frac{5 x}{2}=0, \pi, 2 \pi, 3 \pi, 4 \pi, 5 \pi$
$\Rightarrow x=0, \frac{2 \pi}{5}, \frac{4 \pi}{5}, \frac{6 \pi}{5}, \frac{8 \pi}{5}, 2 \pi$
$\cos \frac{x}{2}=0 \Rightarrow \frac{x}{2}=\frac{\pi}{2} \Rightarrow x=\pi$
$\cos x=0 \Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2}$
So sum $=6 \pi+\pi+2 \pi=9 \pi$
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