Number of values of  $x\in [0,2\pi ]$  and is satisfying the equation $\cos x\cos 2x\cos 3x=1/4$ is  

$2(2\cos 3x\cos x)\cos 2x=1$

$\begin{array}{l}\Rightarrow 2(\cos 4x+\cos 2x)\cos 2x=1\\ \Rightarrow 2\cos 4x\cos 2x=1-2{\cos }^{2}2x=-\cos 4x\\ \Rightarrow \cos 4x(2\cos 2x+1)=0\\ \Rightarrow \cos 4x=0\text{ or }\cos 2x=-1/2\\ \Rightarrow \text{ Now }\cos 4x=0\Rightarrow 4x=(2n+1)\pi /2,n\in \mathbf{I}\\ \Rightarrow x=(2n+1)\pi /8,n\in \mathbf{I}.\\ \Rightarrow \text{ and }0\le (2n+1)\pi /4\le 2\pi \Rightarrow 0\le 2n+1\le 8\\ \Rightarrow n=0,1,2,3.\end{array}$

Also, $\cos \left(2x\right)=-1/2=\cos (2\pi /3)$

$\begin{array}{l}\Rightarrow 2x=2n\pi 2\pi /3,n\in \mathbf{I}\\ \Rightarrow x=n\pi \pi /3,n\in \mathbf{I}.\end{array}$

As $0\le x\le 2\pi ,x=$$\pi /3,2\pi /3,4\pi /3,5\pi /3$

Thus, there are $8$ values of $x.$