Algebra + Trig = fun!

$\begin{aligned} 2 \sin^3 x - 3 \sin x & = - \frac {\color{#3D99F6}3\sin (2x)}2 & \small \color{#3D99F6} \text{Note that }\sin (2x) = 2\sin x \cos x \\ 2 \sin^3 x - 3 \sin x & = - \frac {\color{#3D99F6}6\sin x\cos x}2 & \small \color{#3D99F6} \text{Rearranging} \\ 2 \sin^3 x - 3 \sin x + 3\sin x\cos x & = 0 \\ \sin x (2{\color{#3D99F6}\sin^2 x} - 3 + 3\cos x) & = 0 & \small \color{#3D99F6} \text{Since }\sin^2 x = 1 - \cos^2 x \\ \sin x (-1 - 2\cos^2 x + 3\cos x) & = 0 & \small \color{#3D99F6} \text{Multiply throughout by }-1 \\ \sin x (2\cos^2 x - 3\cos x +1) & = 0 \\ \sin x (\cos x - 1)(2\cos x - 1) & = 0 \end{aligned}$

$\implies \begin{cases} \sin x = 0 & \implies x = 0, \pi, 2 \pi \\ \cos x = 1 & \implies 0, 2\pi \\ \cos x = \dfrac 12 & \implies \dfrac \pi 3, \dfrac {5\pi}3 \end{cases}$

Therefore, the sum of all distinct solutions is $0 + \dfrac \pi 3 + \pi + \dfrac {5\pi}3 + 2\pi = 5\pi$ $\implies a = \boxed 5$ .