<s>[INST] How to solve $\cos x-\sin 3x=\cos 2x$? My attempt:
$$\begin{align} (\cos x- \cos 2x) - \sin 3x &= 0 \\
2\sin \frac{3x}{2}\sin \frac{x}{2} - 2 \sin \frac{3x}{2}\cos \frac{3x}{2}&=0\\
\sin\frac{3x} {2} \left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)&=0 \end{align}$$
Now either $\sin\frac{3x} {2} = 0$ or $\left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)=0$. Solving for the former,
$$\frac{3x}{2}=n\pi\rightarrow x = \frac{2n\pi}{3}$$
How can I solve $\left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)=0$? I know the elementary identities involving trigonometric ratios, but not complex numbers or calculus.
 [/INST] This is an elementary equation $\cos a=\cos b$ under cover: just consider that $\sin a=\cos(\pi/2-a)$. Thus you get
$$
\cos\Bigr(\frac{\pi}{2}-\frac{x}{2}\Bigr)=\cos\frac{3x}{2}
$$
and you get the two families of solutions
$$
\frac{3x}{2}=\frac{\pi}{2}-\frac{x}{2}+2n\pi
\qquad\text{or}\qquad
\frac{3x}{2}=-\frac{\pi}{2}+\frac{x}{2}+2n\pi
$$
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