# 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:$$x=(k+\tfrac14)\pi\quad\text{or}\quad x=(2k-\tfrac12)\pi\quad(k\in\Bbb Z).$$
I think you can use the fact that $$\sin(x) = \cos(\frac{\pi}{2}-x)$$ to translate the equation into one involving only sines (or cosines) on both sides, and then compare directly.
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 $$\frac{3x}{2}=\frac{\pi}{2}-\frac{x}{2}+2n\pi \qquad\text{or}\qquad \frac{3x}{2}=-\frac{\pi}{2}+\frac{x}{2}+2n\pi$$