# Total number of solutions of $\cos x.\cos 2x.\cos 3x=\large\frac{1}{4}$ in $[0,\pi]$ is equal to

$(a)\;4\qquad(b)\;6\qquad(c)\;8\qquad(d)\;10$

$\cos x.\cos 2x.\cos 3x=\large\frac{1}{4}$
$4\cos 3x.\cos x.\cos 2x=1$
$2(\cos 4x+\cos 2x)\cos 2x=1$
$2(2\cos^22x-1+\cos 2x)\cos 2x=1$
$4\cos^32x+2\cos^22x-2\cos 2x-1=0$
$2\cos^22x(2\cos 2x+1)-(2\cos 2x+1)=0$
$(2\cos 2x+1)(2\cos^22x-1)=0$
$\cos 4x(2\cos 2x+1)=0$
$\cos 4x=0$
$4x=(2n_1+1)\large\frac{\pi}{2}$
$x=(2n_1+1)\large\frac{\pi}{8}=\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8}$
$\cos 2x=-\large\frac{1}{2}$
$2x=2n_2\pi\pm \large\frac{2\pi}{3}$
$x=\large\frac{\pi}{3},\frac{2\pi}{3}$
Thus there are 6 solutions.
Hence (b) is the correct answer.