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: $4 \sin x(1 - 2 \sin^2 x)(3 \sin x - 4 \sin^3 x) = 1$
 A: 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?
A: If $\cos{x}=0$ so $x=90^{\circ},$ which is not a root of the equation.
Thus, for $x\neq90^{\circ}$ our equation is equivalent to
$$4\cos{x}\sin{x}\cos2x\sin3x=\cos{x}$$  or
$$\sin4x\sin3x=\cos{x}$$ or
$$\cos{x}-\cos7x=2\cos{x}$$ or
$$\cos7x+\cos{x}=0.$$
Can you end it now?
A: Hint: It is 
$$4\sin(x)\cos(2x)\sin(3x)-1=$$$$-2 \sin \left(\frac{\pi }{4}-2 x\right) \sin \left(2
   x+\frac{\pi }{4}\right) (2 \cos (2 x)-1)$$
A: $$4 \sin(x)\cos(2x)\sin(3x) - 1 = 32 \sin(x)^6 - 40 \sin(x)^4 + 12 \sin(x)^2 - 1$$
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, $$22.5,\; 30,\; 67.5,\; 112.5,\; 150,\; 157.5 $$
