<s>[INST] Solve $\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $ Solve
$$\sin(x) = 4 \sin(60 ^{\circ} -x)\sin(6 ^{\circ} )\sin(18 ^{\circ} ) $$
under degrees (not radians).
There is a well known identity
$$4 \sin x \sin(60 ^{\circ} −x) \sin(60 ^{\circ} +x) = \sin3x$$
but it doesn't quite help here. I can guess a solution $x = 6^{\circ} $, but I would like a more satisfactory solution other than guessing ...
 [/INST] Apply $\sin(60-x)=\frac{\sqrt3}2\cos x -\frac12\sin x$ and
rearrange the equation to isolate $x$
\begin{align}
\cot x &=\frac{1+2 \sin6\sin18} {2\sqrt3\sin 6\sin 18}
\end{align}
Then, substitute $ \sin 18 =\frac{\sin36}{2\cos18}
= \frac{\sin72}{4\cos18\cos36}
= \frac1{4\cos36}$
to obtain
\begin{align}
\cot x &=\frac{2\cos36+\sin6} {\sqrt3\sin 6}
=\frac{\cos36+(\sin54 +\sin6)} {\sqrt3\sin 6}\\
&=\frac{\cos36+\cos24} {\sqrt3\sin 6}
= \frac{2\cos30\cos6} {\sqrt3\sin 6}= \cot 6
\end{align}
Thus, $x = 6^\circ+ 180^\circ n$.
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