$2(\sin x+1)(\sin^2 2x-3\sin x+1) = \sin 4x \cos x$
$\leftrightarrow 2(\sin x+1)(\sin^2 2x-3\sin x+1) = 2\sin 2x \cos 2x\cos x$
$\leftrightarrow (\sin x+1)(\sin^2 2x-3\sin x+1) = 2\sin x \cos^2 x\cos 2x$
$\leftrightarrow \left[\begin{matrix} \sin x+1 = 0 \\ \sin^2 2x-3\sin x+1 = 2\sin x\cos 2x(1-\sin x) \end{matrix}\right.$
$\cdot \ \sin x+1 = 0 \leftrightarrow x = -\dfrac{\pi}{2}+k2\pi \ (k \in \mathbb{Z})$
$\cdot \ \sin^2 2x-3\sin x+1 = 2\sin x\cos 2x(1-\sin x)$
$\leftrightarrow \sin^2 2x-3\sin x+1 = 2\sin x(2\cos^2 x-1)(1-\sin x)$
$\leftrightarrow 4\sin^2 x\cos^2 x-3\sin x+1 = 4\sin x\cos^2 x-4sin^2x\cos^2x-2\sin x+2\sin^2 x$
$\leftrightarrow (2\sin x-1)(4\sin x\cos^2 x-\sin x-1) = 0$
$\leftrightarrow (2\sin x-1)[4\sin x(1-\sin^2 x)-\sin x-1] = 0$
....