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? 

Since
$$ \sin a \cdot \sin b = \frac{1}{2} \left[ \cos (a - b) - \cos (a+b) \right] $$
$$ \sin a \cdot \cos b = \frac{1}{2} \left[ \sin (a + b) + \sin (a - b) \right] $$
and
$$ \sin a - \sin b = 2 \sin\left[ \frac{1}{2} (a-b) \right] \cos \left[ \frac{1}{2} (a+b) \right] $$
we can use these identities to solve
$$ 4 \sin x \cdot \sin 2x \cdot \sin 4x = \sin 3 x $$
$$ \Rightarrow \quad 2 \sin x \cdot \left[ \cos (-2x) - \cos 6x \right] = \sin 3x $$
$$ \Rightarrow \quad 2 \sin x \cdot \cos 2x  - 2 \sin x \cdot \cos 6x = \sin 3x $$
$$ \Rightarrow \quad \sin 3x + \sin (-x)  - (\sin 7x + \sin(-5x)) = \sin 3x $$
$$ \Rightarrow \quad \sin 3x - \sin x  - \sin 7x + \sin 5x = \sin 3x $$
$$ \Rightarrow \quad - \sin x  - \sin 7x + \sin 5x = 0 $$
$$ \Rightarrow \quad \sin x = \sin 5x - \sin 7x $$
$$ \Rightarrow \quad \sin x = 2 \sin (-x) \cos 6x $$
$$ \Rightarrow \quad \sin x = - 2 \sin x \cos 6x $$
$$ \Rightarrow \quad \cos 6x = - \frac{1}{2} $$
$$ \Rightarrow \quad 6x = \frac{2}{3} (3 \pi n \pm \pi) $$
$$ \Rightarrow \quad x = \frac{1}{9} (3 \pi n \pm \pi) $$
or
$$\sin x = 0 \quad \Rightarrow \quad x = n \pi \mathrm{.}$$
