China Mathematical Competition

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 $$-at(t^2-1)-3(t-1)\ge 0,$$ $$(t-1)(-at(t+1)-3)\ge 0.$$ Since $t-1\le 0$, we have $-at(t+1)-3\le 0$, or $$a(t^2+t)\ge -3.①$$ 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, $-\frac{3}{2}\le a\le 12$. ☐