$$\sqrt{3} \cos x - 3 \sin x = 4 \sin 2x \;\cos 3x$$
I tried many things: opening $\sin 2x$, $\cos 3x$, simplifying LHS: $\cos(60^\circ+x)$. Nothing seems to work.
Any hint?
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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
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I tried it, it's making it more complex.. – Harshal Gajjar Jul 20 '14 at 12:59
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2Try WolframAlpha. – Dante Jul 20 '14 at 13:02
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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 $\displaystyle\sin x=\sin\alpha$
lab bhattacharjee
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@ClaudeLeibovici, Thanks. Hope I didn't make any silly mistake. Nice question, I must admit – lab bhattacharjee Jul 20 '14 at 13:17
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