Solve for $x$ where $0\leq x\leq 360$ Solve 
$$4\sin x \cdot \sin 2x \cdot \sin 4x =\sin 3x$$
My Attempt :
Here,
$$4\sin x \cdot  \sin 2x \cdot  \sin 4x=\sin 3x$$
$$4\sin x \cdot (2\sin x  \cdot \cos x ). (4 \sin x  \cdot  \cos x  \cdot \cos 2x)=\sin3x$$
$$32\sin^3 x \cdot \cos^2 x \cdot \cos2x=\sin3x$$
How should I proceed further? 

Using  $\sin3B=3\sin B-4\sin^3B$
$$32\sin^3x\cdot\cos^2x\cdot\cos2x=\sin x(3-4\sin^2x)$$
If $\sin x=0,x=180^\circ n$ where $n$ is any integer
Else using $\cos2A=2\cos^2A-1=1-2\sin^2A,$
$$32\cdot\dfrac{1-\cos2x}2\cdot\dfrac{1+\cos2x}2\cdot\cos2x=3-2(1-\cos2x)$$
$$\iff8(c-c^3)=1+2c\iff4c^3-3c=-\dfrac12$$
$$\implies\cos6x=-\dfrac12,6x=360^\circ m\pm120^\circ\iff x=60^\circ m\pm20^\circ$$ where $m$ is any integer
