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.

We have $\sin\frac12x=\cos\frac32x=\sin(\frac12\pi-\frac32x)$. The general solution to this is$$\tfrac12x=n\pi+(-1)^n(\tfrac12\pi-\tfrac32x)\quad (n\in\Bbb Z).$$The cases of even or odd $n$ then give rise to two sets of solutions:$$x=(k+\tfrac14)\pi\quad\text{or}\quad x=(2k-\tfrac12)\pi\quad(k\in\Bbb Z).$$
