<s>[INST] Solve the equation $4 \sin x \cos (2x) \sin (3x) = 1$ for $0^{\circ} < x < 180^{\circ}$ I have tried using identities but none of them worked for me.  
Eg: $4 \sin x(1 - 2 \sin^2 x)(3 \sin x - 4 \sin^3 x) = 1$
 [/INST] Hint:
Make a partial linearisation: $\;2\sin x\sin 3x=\cos(3x-x)-\cos(3x+x)$, so 
\begin{align}
4 \sin x \cos (2x) \sin (3x)&=2\cos 2x(\cos 2x-\cos 4x)=2\cos^22x-2\cos 2x\cos 4x\\
&=1+\cos 4x-2\cos 2x\cos 4x,
\end{align}
and after simplification, the equation becomes
$$\cos 4x(1-2\cos 2x)=0.$$
Can you solve this one?
</s>