Find the sum of the solutions to
\[2 \sin^3 x - 3 \sin x = -\frac{3}{2} \sin 2x\]in the interval $0 \le x \le 2 \pi.$
By the double-angle formula, $\sin 2x = 2 \sin x \cos x,$ so
\[2 \sin^3 x - 3 \sin x = -3 \sin x \cos x.\]Moving everything to one side, and taking out a factor of $\sin x,$ we get
\[\sin x (2 \sin^2 x - 3 \cos x - 3) = 0.\]From $\sin^2 x = 1 - \cos^2 x,$ $\sin x (2 - 2 \cos^2 x - 3 \cos x - 3) = 0,$ or
\[\sin x (-2 \cos^2 x - 3 \cos x - 1) = 0.\]This factors as
\[-\sin x (\cos x - 1)(2 \cos x - 1) = 0.\]We have that $\sin x = 0$ for $x = 0,$ $\pi,$ and $2 \pi,$ $\cos x = 1$ for $x = 0$ and $x = 2 \pi,$ and $\cos x = \frac{1}{2}$ for $x = \frac{\pi}{3}$ and $x = \frac{5 \pi}{3}.$  Thus, the sum of the solutions is
\[0 + \frac{\pi}{3} + \pi + \frac{5 \pi}{3} + 2 \pi = \boxed{5 \pi}.\]