Question

Solve $\cos 4x={\cos }^{2}3x$.

Answer 1

Change in terms of $2x$.
$2{\cos }^{2}2x-1=\frac{1}{2}(1+\cos 6x)$
$\therefore 4c^2-2=1+4c^3-3c$, where $c=\cos 2x$
or $4c^3-4c^2-3c+3=0$
or $(c-1)(4c^2-3)=0$
$\therefore \cos 2x=1$, $\cos 2x=\pm \frac{\sqrt{3}}{2}=\pm \cos \frac{\pi }{6}$
$\therefore 2x=2n\pi$ or $2n\pi \pm \left(\pi /6\right)$
or $x=n\pi$ or $n\pi \pm \left(\pi /12\right)$.