The number of solutions of $$\sin^2 x + (2 + 2x - x^2)\sin x - 3(x - 1)^2 = 0$$, where $$-\pi \leq x \leq \pi$$, is ________

We wish to find the number of solutions of $$\sin^2 x + (2+2x-x^2)\sin x - 3(x-1)^2 = 0$$ for $$-\pi \leq x \leq \pi$$.

Letting $$s = \sin x$$ and observing that $$2+2x-x^2 = 3-(x-1)^2$$ transforms the equation into $$s^2 + \bigl(3-(x-1)^2\bigr)s - 3(x-1)^2 = 0,$$ which can be rewritten as $$s^2 + 3s - (x-1)^2(s+3) = 0$$ and thus factors to $$(s+3)(s-(x-1)^2) = 0.$$

It follows that $$\sin x + 3 = 0$$, giving $$\sin x = -3$$, which is impossible since $$|\sin x|\leq1$$. Therefore we must have $$\sin x = (x-1)^2$$, requiring $$0 \leq (x-1)^2 \leq 1$$ and hence $$x \in [0,2].$$

On the interval $$[0,2]$$, comparing $$y = \sin x$$ and $$y = (x-1)^2$$ shows that at $$x = 0$$, $$\sin0 = 0$$ is below $$(0-1)^2 = 1$$; at $$x = 1$$, $$\sin1 \approx 0.841$$ is above $$(1-1)^2 = 0$$; and at $$x = 2$$, $$\sin2 \approx 0.909$$ is below $$(2-1)^2 = 1$$. Thus the two curves intersect twice, once in $$(0,1)$$ and once in $$(1,2)$$.

Therefore there are 2 solutions, and the correct answer is $$\boxed{2}$$.