Question

Solve:
$\cos 6x=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$

Answer 1

$\cos 6x=\cos 2\left(3x\right)=2{\cos }^{2}3x-1$

Also, $\cos 3x=4{\cos }^{3}x-3\cos x$

$\cos 6x=2{\cos }^{2}3x-1$

$=2{(4{\cos }^{3}x-3\cos x)}^2-1$

$=2(16{\cos }^{6}x+9{\cos }^{2}x-24{\cos }^{4}x)-1$

$=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$

Hence proved.