$\sin 4x +\sqrt{3} \sin 3 x + \sin 2 x=0$ This question is from a 2012 VMK entrance exam
I was trying to solve it first by expanding $\sin 4 x = 2 \sin 2 x \cos 2x$, then by noticing that if divided by 2, one can get, e.g. $ \frac{\sqrt{3}}{2} \cos 3 x  = \sin \frac{\pi}{6} \cos 3 x$ and then trying to find some way to obtain $\sin (a +b)$ or $\cos (a+b)$, but without any success. 
I suspect there should be some adding an extra term and then multiplying out, but can't get my head around it.

If you don't see any trick, you can always try
\begin{align*}
&\sin(4x)=\text{Im}(\cos x+i\sin x)^4=4\cos^3\!x\sin x-4\cos x\sin^3\!x\\
&\sin(2x)=\text{Im}(\cos x+i\sin x)^3=3\cos^2\!x\sin x-\sin^3\!x\\
&\sin(3x)=\text{Im}(\cos x+i\sin x)^2=2\cos x\sin x.
\end{align*}
Plugging it in yields
$$(4\cos^3\!x-4\cos x\sin^2\!x+\sqrt3(3\cos^2\!x-\sin^2\!x)+2\cos x)\cdot\sin x=0$$
and you can solve the left factor by setting $y=\cos x$
\begin{align*}
0&=4y^3-4y(1-y^2)+\sqrt3(3y^2-(1-y^2))+2y=8y^3-4y+\sqrt3(4y^2-1)+2y\\
&=2y(4y^2-1)+\sqrt3(4y^2-1)=(2y+\sqrt3)(2y+1)(2y-1).
\end{align*}
I'm sure you could finish this.
