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

$$4 \sin(x)\cos(2x)\sin(3x) - 1 = 32 \sin(x)^6 - 40 \sin(x)^4 + 12 \sin(x)^2 - 1$$
Letting $\sin(x) = s$, this is 
$$ 32 s^6 - 40 s^4 + 12 s^2 - 1 = (2s+1)(2s-1)(8s^4 - 8s^2+1)$$
which has roots 
$$ s = \pm \frac{1}{2},\; \pm \frac{\sqrt{2\pm\sqrt{2}}}{2} $$
The values of $x$ in $(0,\pi)$ are $$\frac{\pi}{8},\; \frac{\pi}{6},\; \frac{3\pi}{8},\; \frac{5\pi}{8}\; \frac{5\pi}{6},\; \frac{7\pi}{8}$$
or in degrees, $$22.5,\; 30,\; 67.5,\; 112.5,\; 150,\; 157.5 $$
