Solve the following equations
$\cos x-\cos 2x+\cos 3x=0$.

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

Use :

$\cos 2x=2{\cos }^{2}x-1$

$\cos 3x=4{\cos }^{3}x-3\cos x$

Now,

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

$\cos x-(2{\cos }^{2}x-1)+(4{\cos }^{3}x-3\cos x)=0$

$4{\cos }^{3}x-2{\cos }^{2}x-2\cos x+1=0$

$2{\cos }^{2}x[2\cos x-10-1[2\cos x-1]=0$

$(2\cos x-1)(2{\cos }^{2}x-1)=0$

$\Rightarrow \cos x=\frac{1}{2}\Rightarrow =\pm \frac{\pi }{3}+2\pi n$

$\Rightarrow {\cos }^{2}x=\frac{1}{2}\Rightarrow \cos x=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow x=\frac{\pi }{4}+2\pi n$.