Question

Solve the following equation:
$\cos 2x-\cos 8x+\cos 6x=1$

Answer 1

$\cos 2x+\cos 6x-\cos 8x=1$

$2\cos 4x\cos 2x=1+\cos 8x={2\cos }^{2}4x$

$⟹\cos 4x=0$

$4x=(2n+1)\frac{\pi }{2}$

$x=(2n+1)\frac{\pi }{8}.$

And $⟹\cos 2x=\cos 4x$

$\cos 4x-\cos 2x=0$
$-2\sin x\sin x=0$

$\sin 3x=0,3x=n\pi$

$x=\frac{n\pi }{3}.$

or $\sin x=0$

$x=n\pi .$