Solving the equation: $3\cos x - \sin 2x = \sqrt{3}(\cos 2x + \sin x)$ 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!
 A: Hint
$$\cos{x}(3-2\sin{x})=\sqrt{3}(2\sin{x}+1)(1-\sin{x})$$
$$\Longrightarrow \left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)\left(\cos{\dfrac{x}{2}}+\sin{\dfrac{x}{2}}\right)(3-2\sin{x})=\sqrt{3}\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)^2(2\sin{x}+1)$$
(1):$$\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)=0\Longrightarrow x=\dfrac{\pi}{2}+2k\pi$$
(2):$$\left(\cos{\dfrac{x}{2}}+\sin{\dfrac{x}{2}}\right)(3-2\sin{x})=\sqrt{3}\left(\cos{\dfrac{x}{2}}-\sin{\dfrac{x}{2}}\right)(2\sin{x}+1)\tag{1}$$
let
$$\sin{\dfrac{x}{2}}+\cos{\dfrac{x}{2}}=t\Longrightarrow \sin{x}=t^2-1,\left(\sin{\dfrac{x}{2}}-\cos{\dfrac{x}{2}}\right)^2=2-t^2$$
use $(1)^2$
then we have
$$t^2\cdot(3-2t^2+2)^2=3(2-t^2)(2t^2-1)^2\Longrightarrow 16t^6-56t^4+52t^2-6=0$$
$$\Longrightarrow (2t^2-3)(2t^2-2t-1)(2t^2+2t-1)=0$$
so
$$\Longrightarrow  t=\sqrt{\dfrac{3}{2}},-\sqrt{\dfrac{3}{2}},\dfrac{1}{2}(-1\pm \sqrt{3})-\dfrac{1}{2}(1+\sqrt{3})$$
A: 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
$$\sqrt3\sin(\frac\pi3-x)=\sin(\frac{2\pi}3-2x)$$
$$\sqrt3\sin(\frac\pi3-x)=2\sin(\frac\pi3-x)\cos(\frac\pi3-x)$$
$$\sin(\frac\pi3-x)[\sqrt3-2\cos(\frac\pi3-x)]=0$$
