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

Using $\cos 3x = \cos^3 x -3\cos x \sin^2 x$,
$$\begin{align*}
\cos 3x &= \cos^2 x - 3 \sin ^2 x\\
\cos^3 x -3\cos x \sin^2 x &= \cos^2 x - 3 \sin ^2 x\\
(\cos x)(\cos^2 x - 3 \sin ^2 x) &= \cos^2 x - 3 \sin ^2 x\\
\cos x &= 1&\text{ or }&& \cos^2 x - 3 \sin ^2 x &= 0\\
\cos x &= 1&\text{ or }&&\tan^2 x&=\frac13
\end{align*}$$

My formula of $\cos 3x$ may look non-standard, but that was because I did not remember it and instead got it by quickly expanding
$$\cos 3x + i\sin 3x = (\cos x + i\sin x)^3$$
