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: $4 \sin x(1 - 2 \sin^2 x)(3 \sin x - 4 \sin^3 x) = 1$

Hint: It is 
$$4\sin(x)\cos(2x)\sin(3x)-1=$$$$-2 \sin \left(\frac{\pi }{4}-2 x\right) \sin \left(2
   x+\frac{\pi }{4}\right) (2 \cos (2 x)-1)$$
