The equation  $8\cos x\cos 2x\cos 4x=\frac{\sin 6x}{\sin x}$has a solution given by

$8\cos x\cos 2x\cos 4x=\frac{\sin 6x}{\sin x}$

$\sin 8x=\sin 6x$

$\sin 8x-\sin 6x=0$

$2\cos 7x\sin x=0$

$7x=\{(2n+1)\frac{\pi }{2};n\in Z\}\text{ or }x=\{n\pi ;n\in Z\}$

$x=\left\{(2n+1)\frac{\pi }{14}\right\}\text{ or }x=\{n\pi ;n\in Z\}$

at $n\pi ;n\in Z$ question not defined