<s>[INST] Trigonometric Equation : $\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$ Please help solving this equation: 
$\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)$
I used numerical method to solve it and got $x=6^\circ$ but I am not able  to solve it by trigonmetry.
Thank!
Best Regards, Michael.
 [/INST] Using the same observation as @dantopa we get
\begin{eqnarray}
\sin(x) &=& \sin(12^\circ - x )\\
\sin(x) &=& \sin(12^\circ)\cos(x)-\cos(12^\circ)\sin(x)\\
\tan(x) &=& \sin(12^\circ)-\cos(12^\circ)\tan(x)\\
\tan(x)(1+\cos(12^\circ))&=&\sin(12^\circ)\\
\tan(x)&=&\frac{\sin(12^\circ)}{1+\cos(12^\circ)}\\
\tan(x)&=&\tan(6^\circ)\\
x &=& 6^\circ+180^\circ n
\end{eqnarray}
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