$\text{The total number of solutions of the equation}\cos x.\cos 2x.\cos 3x=\frac{1}{4}\text{in}[0,\pi ]\text{is}$

The correct option is B

$6$

Given
$\cos x.\cos 2x.\cos 3x=\frac{1}{4}$
$\Rightarrow 4(\cos 3x.\cos x)\cos 2x=1$
$\Rightarrow 2(2\cos 3x.\cos x)\cos 2x=1$
$\Rightarrow 2(\cos 4x+\cos 2x)\cos 2x=1$
$\Rightarrow 2(2{\cos }^{2}2x-1+\cos 2x)\cos 2x=1$
$\Rightarrow 4{\cos }^{3}2x+2{\cos }^{2}2x-2\cos x-1=0$
$\Rightarrow 2{\cos }^{2}2x(2\cos 2x+1)-(2\cos 2x+1)=0$
$\Rightarrow (2{\cos }^{2}2x-1)(2\cos 2x+1)=0$
$\Rightarrow \cos 4x.(2\cos 2x+1)=0$
$\cos 4x=0\text{or}2\cos 2x+1=0$
$\cos 4x=\cos \frac{\pi }{2}\text{or}\cos 2x=\frac{-1}{2}$
$4x=(2n_1+1)\frac{\pi }{2},n_1\in Z\text{or}2x=2n_2\pi \pm \frac{2\pi }{3},n_2\in Z$
$x=(2n_1+1)\frac{\pi }{8},n_1\in Z\text{or}x=n_2\pi \pm \frac{\pi }{3},n_2\in Z$
$\therefore x=\frac{\pi }{8},\frac{3\pi }{8},\frac{5\pi }{7},\frac{7\pi }{8}\text{or}x=\frac{\pi }{3},\frac{2\pi }{3}$

$\therefore$ The equation has 6 solutions as $x\in [0,\pi ]$

Hence the correct answer is Option B.