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

$\cos 4 x=\cos 2 x$

$\Rightarrow \cos 4 x-\cos 2 x=0$

$\Rightarrow-2 \sin \left(\frac{4 x+2 x}{2}\right) \sin \left(\frac{4 x-2 x}{2}\right)=0$

$\left[\because \cos \mathrm{A}-\cos \mathrm{B}=-2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\right]$

$\Rightarrow \sin 3 x \sin x=0$

$\Rightarrow \sin 3 x=0 \quad$ or $\quad \sin x=0$

$\therefore 3 \mathrm{x}=\mathrm{n} \pi \quad$ or $\quad \mathrm{x}=\mathrm{n} \pi$, where $\mathrm{n} \in \mathrm{Z}$

$\Rightarrow x=\frac{n \pi}{3} \quad$ or $\quad x=n \pi$, where $n \in Z$