Let $\sin x = t$. The expression is then changed to
$$g(t) = (-a t^2 + a - 3)t,$$
or
$$g(t) = -a t^3 + (a-3)t.$$
From $-a t^3 + (a-3)t \ge -3$, we get
$$-a t (t^2 - 1) - 3 (t - 1) \ge 0,$$
$$(t - 1)(-a t (t + 1) - 3) \ge 0.$$
Since $t - 1 \le 0$, we have $-a t (t + 1) - 3 \le 0$, or
$$a (t^2 + t) \ge -3. \qquad ①$$
When $t = 0, -1$, expression ① always holds; when $0 < t \le 1$, we have $0 < t^2 + t \le 2$; and when $-1 < t < 0$, $-\frac{1}{4} \le t^2 + t < 0$. Therefore, $-\frac{3}{2} \le a \le 12$. ☐