Find the general solution of the equation $cos 3 x+cos x-cos 2 x=0$

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

$\Rightarrow 2 \cos \left(\frac{3 x+x}{2}\right) \cos \left(\frac{3 x-x}{2}\right)-\cos 2 x=0 \quad\left[\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

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

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

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

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

$\therefore 2 \mathrm{x}=(2 \mathrm{n}+1) \frac{\pi}{2} \quad$ or $\quad \cos \mathrm{x}=\cos \frac{\pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$

$\Rightarrow \mathrm{x}=(2 \mathrm{n}+1) \frac{\pi}{4} \quad$ or $\quad \mathrm{x}=2 \mathrm{n} \pi \pm \frac{\pi}{3}$, where $\mathrm{n} \in \mathrm{Z}$