<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] Observation
$$
 \sin 96^{\circ} \sin 12^{\circ} =  \sin 18^{\circ} \sin 42^{\circ} 
$$
Implication
$$
 \sin x = \sin \left(12^{\circ} - x \right) \qquad \Rightarrow \qquad x = 12^{\circ} - x 
$$
Conclusion

$$x=6^{\circ}$$

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