Question

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

• A. $8\pi$
• B. $11\pi$
• C. $12\pi$
• D. $9\pi$

Answer 1

Answer: (D)

$(\sin x+\sin 4x)+(\sin 2x+\sin 3x)=0$
$\Rightarrow 2\sin \frac{5x}{2}\left\{\cos \frac{3x}{2}+\cos \frac{x}{2}\right\}=0$
$\Rightarrow 2\sin \frac{5x}{2}\left\{2\cos x\cos \frac{x}{2}\right\}=0$
$2\sin \frac{5x}{2}=0\Rightarrow \frac{5X}{2}=0,\pi ,2\pi ,3\pi ,4\pi ,5\pi$
$\Rightarrow x=0,\frac{2\pi }{5},\frac{4\pi }{5},\frac{6\pi }{5},\frac{8\pi }{5},2\pi$
$\cos \frac{x}{2}=0\Rightarrow \frac{x}{2}=\frac{\pi }{2}\Rightarrow x=\pi$
$\cos x=0\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2}$
So Sum $=6\pi +\pi +2\pi =9\pi$