TITLE: Solving the equation: $3\cos x - \sin 2x = \sqrt{3}(\cos 2x + \sin x)$
QUESTION [7 upvotes]: Solving the equation:
$$3\cos x - \sin 2x = \sqrt{3}(\cos 2x + \sin x)\tag{1}$$
I tried to write $(1)$ becomes
$$\sqrt{3}\sin \left(\frac{\pi}{3}-x\right)=\sin \left(\frac{\pi}{3}+2x \right)$$
Now, I have stuck :( , can you help me?
Thanks!

REPLY [2 votes]: After staring at this for a while I think I've got it.  Assuming what you've done so far is correct
$$\sin(\frac\pi3+2x)=\sin[\pi-(\frac\pi3+2x)]=\sin(\frac{2\pi}3-2x)$$
Substituting this gives
$$\sqrt3\sin(\frac\pi3-x)=\sin(\frac{2\pi}3-2x)$$
$$\sqrt3\sin(\frac\pi3-x)=2\sin(\frac\pi3-x)\cos(\frac\pi3-x)$$
$$\sin(\frac\pi3-x)[\sqrt3-2\cos(\frac\pi3-x)]=0$$