Solve $\sin 2 x-\sin 4 x+\sin 6 x=0$
Solve $\sin 2 x-\sin 4 x+\sin 6 x=0$
The equation can be written as
$\sin 6 x+\sin 2 x-\sin 4 x=0$
or $2 \sin 4 x \cos 2 x-\sin 4 x=0$
i.e. $\quad \sin 4 x(2 \cos 2 x-1)=0$
Therefore $\sin 4 x=0 \quad$ or $\cos 2 x=\frac{1}{2}$
i.e. $\sin 4 x=0$ or $\cos 2 x=\cos \frac{\pi}{3}$
Hence $\quad 4 x=n \pi$ or $2 x=2 n \pi \pm \frac{\pi}{3},$ where $n \in Z$
i.e. $x=\frac{n \pi}{4}$ or $x=n \pi \pm \frac{\pi}{6},$ where $n \in Z$
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- [IIT 2013]