Question

$8\cos x\cos 2x\cos 4x=\frac{\sin 6x}{\sin x}$ Find the value of $x$.

Answer 1

Multiplying by $\sin x(\sin x\ne 0)$, we get
$\sin 8x-\sin 6x=0$
$2\sin x\cos 7x=0$
$\therefore \cos 7x=0$ as $\sin x\ne 0$
$\therefore 7x=(n+\frac{1}{2})\pi$
$\therefore x=(n+\frac{1}{2})\frac{\pi }{7}$.