help with trigonometric equations How do I solve this?
$$\cos3x=\cos^2x-3\sin^2x$$

$$ \cos(3x)=4\cos^3(x)-3\cos(x),\quad 4\cos^2(x)-3=\cos^2 x-3\sin^2 x $$
hence by setting $z=\cos x$ we have to solve:
$$ 4z^3-4z^2-3z+3 = (z-1)(4z^2-3) = 0 $$
so $\cos x=1$ or $\cos x=\frac{\sqrt{3}}{2}$, from which $x=2k\pi$ or $x=\pm\frac{\pi}{6}+2k\pi$.
