<s>[INST] $\sin 4x +\sqrt{3} \sin 3 x + \sin 2 x=0$ 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.
 [/INST] 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.
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