The number of solutions of the equation $$4\sin^2 x - 4\cos^3 x + 9 - 4\cos x = 0$$; $$x \in [-2\pi, 2\pi]$$ is:

We need to find the number of solutions of $$4\sin^2 x - 4\cos^3 x + 9 - 4\cos x = 0$$ in $$[-2\pi, 2\pi]$$.

$$4(1 - \cos^2 x) - 4\cos^3 x + 9 - 4\cos x = 0$$

$$4 - 4\cos^2 x - 4\cos^3 x + 9 - 4\cos x = 0$$

$$-4\cos^3 x - 4\cos^2 x - 4\cos x + 13 = 0$$

$$4\cos^3 x + 4\cos^2 x + 4\cos x - 13 = 0$$

$$4t^3 + 4t^2 + 4t - 13 = 0$$

Let $$g(t) = 4t^3 + 4t^2 + 4t - 13$$.

Check $$g(1) = 4 + 4 + 4 - 13 = -1 < 0$$.

Check $$g(-1) = -4 + 4 - 4 - 13 = -17 < 0$$.

Since $$g(t)$$ is a cubic with positive leading coefficient, it goes from $$-\infty$$ to $$+\infty$$. But we need to check if it has any root in $$[-1, 1]$$.

$$g'(t) = 12t^2 + 8t + 4$$. The discriminant is $$64 - 192 = -128 < 0$$, so $$g'(t) > 0$$ for all $$t$$. This means $$g(t)$$ is strictly increasing.

Since $$g$$ is strictly increasing and $$g(1) = -1 < 0$$, we have $$g(t) < 0$$ for all $$t \leq 1$$.

Therefore, there is no value of $$t \in [-1, 1]$$ satisfying the equation.

Since the equation has no solution for $$\cos x \in [-1, 1]$$, there are 0 solutions in $$[-2\pi, 2\pi]$$.

The correct answer is 0 (Option D).