Find the general solution of the equation $\cos 4x=\cos 2x$

Step 1: Simplification
Given: $\cos 4x=\cos 2x$
$\Rightarrow \cos 4x-\cos 2x=0$

$\Rightarrow -2\sin \left(\frac{4x+2x}{2}\right)\sin \left(\frac{4x-2x}{2}\right)=0[\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)]$
$\Rightarrow -2\sin \left(\frac{6x}{2}\right)\sin \left(\frac{2x}{2}\right)=0$
$\Rightarrow -2\sin 3x\sin x=0$
$\Rightarrow \sin 3x\sin x=0$
So, either $\sin 3x=0$ or $\sin x=0$
We solve $\sin 3x=0$ and $\sin x=0$ separately.

Step 2: General solution for $\sin 3x=0$
$\sin 3x=0$ or $\sin 3x=\sin 0$
So the general solution is
$3x=n\pi \pm (-1{)}^n(0)$
$\Rightarrow 3x=n\pi$
$\Rightarrow x=\frac{n\pi }{3}$

Step 3: General solution for $\sin x=0$
$\sin x=0$
$\Rightarrow x=n\pi$

Final answer $:x=n\pi ,\frac{n\pi }{3}$