Answer
$x=\dfrac{\pi}{4}, \dfrac{3\pi}{4},\dfrac{4\pi}{3}, \dfrac{2\pi}{3} $
Work Step by Step
Given: $\cos 3x+\cos 2x+\cos x=0$
This gives:
$2 \cos (\dfrac{3x+x}{2})\cos(\dfrac{3x-x}{2})+\cos 2x=0$
or, $2 \cos (\dfrac{4x}{2})\cos(\dfrac{2x}{2})+\cos 2x=0$
or, $ 2 \cos 2x\cos x+\cos 2x=0$
or, $\cos 2x[1+2 \cos x]=0$
Either $\cos 2x=0$ or, $1+2 \cos x=0$
$\cos 2x=0 \implies x=\dfrac{\pi}{4}, \dfrac{3\pi}{4}$
or, $ \cos x=\dfrac{-1}{2} \implies x=\dfrac{4\pi}{3}, \dfrac{2\pi}{3} $
