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 $$[-2\pi, 2\pi]$$.

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

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

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

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

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

We know that, $$\cos x \leq 1$$

$$\Rightarrow$$ $$4\cos^3 x \leq 4$$

$$\Rightarrow$$ $$4\cos^2 x \leq 4$$

$$\Rightarrow$$ $$4\cos x \leq 4$$

$$\therefore$$ $$4\cos^3 x + 4\cos^2 x + 4\cos x \leq 12$$

Thus, the maximum value of LHS can be $$12$$, and RHS = $$13$$

Thus, there is no solution for the equation between $$x \in [-2\pi, 2\pi]$$

Hence, option D is the correct choice.