Prove $\sin2x+\sin4x+\sin6x=4\cos x\cos2x\sin3x$ Prove $\sin{2x}+\sin{4x}+\sin{6x}=4\cos{x}\cos{2x}\sin{3x}$
I have reached the point where the LHS equation has turned into $2\cos{x}\cos2x\sin{x}(2\sin2x+1)$
But I have no idea how to turn $\sin{x}(2\sin2x+1)$ into $2\sin3x$
A quicker method if it exists would be greatly appreciated
Thanks in advance

To solve this problem, we can use the identities:
$$
\sin A + \sin B = 2\sin \frac{A+B}{2} \cos \frac{A - B}{2},
$$
$$
\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A - B}{2},
$$
and
$$
\sin 2\phi = 2\sin \phi \cos \phi.
$$
Going back to the question,
$
\text{LHS} = \sin 2x + \sin 4x + \sin 6x \\
= 2\sin 3x \cos x + \sin 6x \\
= 2\sin 3x \cos x + 2\sin 3x \cos 3x \\
= 2\sin 3x (\cos x + \cos 3x) \\
= 2\sin 3x \times 2\cos 2x \cos x \\
= 4\cos x \cos 2x \sin 3x \\
= \text{RHS}.
$
Hence, proved.
