The number of roots in $[0,\mathrm{\pi }]$ of the equation  $\cos \theta \cos 2\theta \cos 3\theta =\frac{1}{4}$ is

$4\cos \theta \cos 2\theta \cos 3\theta =1$

$\Rightarrow 2\cos \theta \cos 3\theta \cdot 2\cos 2\theta =1$

$\begin{array}{l}\Rightarrow 2(\cos 4\theta +\cos 2\theta )\cos 2\theta =1\\ \Rightarrow 2\cos 4\theta \cos 2\theta +2{\cos }^{2}2\theta -1=0\\ \Rightarrow \cos 4\theta (2\cos 2\theta +1)=0\\ \therefore \cos 4\theta =0\Rightarrow \theta =\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{8},\frac{7\pi }{8}\\ 2\cos 2\theta +1=0\Rightarrow \cos 2\theta =-\frac{1}{2}\therefore \theta =\frac{\pi }{3},\frac{2\pi }{3}.\end{array}$