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

The given equation is $\sin 6x + \sin 2x - \sin 4x = 0$.
Using the sum-to-product formula $\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$,we get:
$2 \sin 4x \cos 2x - \sin 4x = 0$.
Factoring out $\sin 4x$:
$\sin 4x (2 \cos 2x - 1) = 0$.
This implies either $\sin 4x = 0$ or $\cos 2x = \frac{1}{2}$.
For $\sin 4x = 0$,the general solution is $4x = n\pi$,which gives $x = \frac{n\pi}{4}$ for $n \in \mathbb{Z}$.
For $\cos 2x = \frac{1}{2} = \cos \frac{\pi}{3}$,the general solution is $2x = 2n\pi \pm \frac{\pi}{3}$,which gives $x = n\pi \pm \frac{\pi}{6}$ for $n \in \mathbb{Z}$.