$\cos 3 x+\cos x-\cos 2 x=0$

$ \Rightarrow 2\cos \left( {\frac{{3x + 2}}{2}} \right)\cos \left( {\frac{{3x – x}}{2}} \right) – \cos 2x = 0\quad $

$\left[ {\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A – B}}{2}} \right)} \right]$

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

$\Rightarrow \cos 2 x(2 \cos x-1)=0$

$\Rightarrow \cos 2 x=0 \quad$ or $\quad 2 \cos x-1=0$

$\Rightarrow \cos 2 x=0 \quad$ or $\quad \cos x=\frac{1}{2}$

$\therefore 2 x=(2 n+1) \frac{\pi}{2}$

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

$\Rightarrow x=(2 n+1) \frac{\pi}{4}$

or $\quad x=2 n \pi \pm \frac{\pi}{3},$ where $n \in Z$