help with trigonometric equations How do I solve this?
$$\cos3x=\cos^2x-3\sin^2x$$
 A: If you write $\cos(3x) = \cos(x+2x)$, you can then apply the identity for the cosine of a sum.  After that you can apply the identities that reduce $\cos(2x)$ and $\sin(2x)$ to functions of $\cos x$ and $\sin x$.  You'll end up with an expression in which all sines are squared, so you can apply the identity $\sin^2 x = 1-\cos^2 x$.  You're left with no trigonometric expressions except $\cos x$.  Then the substitution $u=\cos x$ reduces it to an algebraic equation to be solved for $u$.
You shouldn't wait until you know how to solve it before you start working on it.  If you start with $\cos(x+2x)$ and apply the identity for the cosine of a sum, you can just see where that takes you.  That's how I did this.
A: $$ \cos(3x)=4\cos^3(x)-3\cos(x),\quad 4\cos^2(x)-3=\cos^2 x-3\sin^2 x $$
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 $x=\pm\frac{\pi}{6}+2k\pi$.
A: Hint: use $\cos(3x)=4\cos(x)^3-3\cos(x)$ and $\sin(x)^2=1-\cos(x)^2$
A: 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
$$\cos 3x + i\sin 3x = (\cos x + i\sin x)^3$$
