Question

Solve $\sin 2x-\sin 4x+\sin 6x=0$

Answer 1

Step 1: Simplification
Given $\sin 2x-\sin 4x+\sin 6x=0$
$\Rightarrow (\sin 2x+\sin 6x)-\sin 4x=0$
$\Rightarrow 2\sin \frac{6x+2x}{2}\cos \frac{6x-2x}{2}-\sin 4x=0$
$\Rightarrow 2\sin \frac{8x}{2}\cos \frac{4x}{2}-\sin 4x=0$
$\Rightarrow 2\sin 4x\cos 2x-\sin 4x=0$
$\Rightarrow \sin 4x(2\cos 2x-1)=0$
$\Rightarrow \sin 4x=0,2\cos 2x=1$

We need to find general solution for both separately.
Step 2: General solution of $\sin 4x=0$
General solution is $4x=n\pi \Rightarrow x=\frac{n\pi }{4}$ where $n\in z$

Step 3: General solution of $\cos 2x=\frac{1}{2}$
$\Rightarrow \cos 2x=\cos \left(\frac{\pi }{3}\right)$
$\therefore 2x=2n\pi \pm \frac{\pi }{3}\Rightarrow x=n\pi \pm \frac{\pi }{6}$ where $n\in z$