# Solving this trigonometric equation

$$\sqrt{3} \cos x - 3 \sin x = 4 \sin 2x \;\cos 3x$$

I tried many things: opening $\sin 2x$, $\cos 3x$, simplifying LHS: $\cos(60^\circ+x)$. Nothing seems to work.

Any hint?

-
Try using $\sin 2x = 2\sin x \cos x$, $\cos 3x = \cos 2x \cos x - \sin 2x \sin x$ and $\cos 2x = \cos^2 x - \sin^ x$ –  Darth Geek Jul 20 '14 at 12:54
I tried it, it's making it more complex.. –  Harshal Gajjar Jul 20 '14 at 12:59
Try WolframAlpha. –  user96402 Jul 20 '14 at 13:02

Werner Formula says: $$2\sin2x\cos3x=\sin5x-\sin x$$

So, we have $$\sqrt3\cos x-3\sin x=2(\sin5x-\sin x)$$

$$\iff\sqrt3\cos x-\sin x=2\sin5x$$

$$\iff2\sin(60^\circ -x)=2\sin5x$$

Hope you know about the general solution of $\displaystyle\sin x=\sin\alpha$

-
So simple and smart ! –  Claude Leibovici Jul 20 '14 at 13:13
@ClaudeLeibovici, Thanks. Hope I didn't make any silly mistake. Nice question, I must admit –  lab bhattacharjee Jul 20 '14 at 13:17
Thanks Lab bhattacharjee, you're always ready to help! :) –  Harshal Gajjar Jul 25 '14 at 13:22