Question

Solve the following equations.
$2\cos 4x-\cos 2x=4{\cos }^{2}x-1.$

Answer 1

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

${4\cos }^{2}2x-3\cos 2x-3=0$

Let, $\cos 2x=z.$

${4z}^2-3z-3=0$

$z=\frac{3\pm \sqrt{9+48}}{8}=\frac{3\pm \sqrt{57}}{8}.$

Let, $a=\frac{3\pm \sqrt{57}}{8}.$

So, $2x=2n\pi +{\cos }^{-1}\left(a\right),x=n\pi \pm \frac{1}{2}{\cos }^{-1}\left(a\right).$