The number of solutions of equation $$(4 - \sqrt{3})\sin x - 2\sqrt{3}\cos^2 x = \dfrac{-4}{1 + \sqrt{3}}$$, $$x \in \left[-2\pi, \dfrac{5\pi}{2}\right]$$ is

The equation is $$(4 - \sqrt{3})\sin x - 2\sqrt{3}\cos^2 x = \dfrac{-4}{1 + \sqrt{3}}$$.

Rationalizing the RHS: $$\dfrac{-4}{1 + \sqrt{3}} \cdot \dfrac{1 - \sqrt{3}}{1 - \sqrt{3}} = \dfrac{-4(1 - \sqrt{3})}{1 - 3} = \dfrac{-4(1 - \sqrt{3})}{-2} = 2(1 - \sqrt{3}) = 2 - 2\sqrt{3}$$.

Using $$\cos^2 x = 1 - \sin^2 x$$:

$$(4 - \sqrt{3})\sin x - 2\sqrt{3}(1 - \sin^2 x) = 2 - 2\sqrt{3}$$

$$(4 - \sqrt{3})\sin x - 2\sqrt{3} + 2\sqrt{3}\sin^2 x = 2 - 2\sqrt{3}$$

$$2\sqrt{3}\sin^2 x + (4 - \sqrt{3})\sin x - 2 = 0$$

Let $$t = \sin x$$. Using the quadratic formula:

$$t = \dfrac{-(4 - \sqrt{3}) \pm \sqrt{(4 - \sqrt{3})^2 + 16\sqrt{3}}}{4\sqrt{3}}$$

$$(4 - \sqrt{3})^2 = 16 - 8\sqrt{3} + 3 = 19 - 8\sqrt{3}$$

$$19 - 8\sqrt{3} + 16\sqrt{3} = 19 + 8\sqrt{3} = (4 + \sqrt{3})^2$$

So $$t = \dfrac{-(4 - \sqrt{3}) \pm (4 + \sqrt{3})}{4\sqrt{3}}$$

Taking the positive sign: $$t = \dfrac{2\sqrt{3}}{4\sqrt{3}} = \dfrac{1}{2}$$

Taking the negative sign: $$t = \dfrac{-8}{4\sqrt{3}} = \dfrac{-2}{\sqrt{3}} = \dfrac{-2\sqrt{3}}{3}$$. Since $$|\sin x| \le 1$$ and $$2\sqrt{3}/3 \approx 1.155 \gt 1$$, this is rejected.

So $$\sin x = \dfrac{1}{2}$$, giving $$x = \dfrac{\pi}{6} + 2n\pi$$ or $$x = \dfrac{5\pi}{6} + 2n\pi$$.

In $$\left[-2\pi, \dfrac{5\pi}{2}\right]$$:

For $$x = \dfrac{\pi}{6} + 2n\pi$$: $$n = -1$$ gives $$x = -\dfrac{11\pi}{6}$$; $$n = 0$$ gives $$x = \dfrac{\pi}{6}$$; $$n = 1$$ gives $$x = \dfrac{13\pi}{6}$$. All three are in the range.

For $$x = \dfrac{5\pi}{6} + 2n\pi$$: $$n = -1$$ gives $$x = -\dfrac{7\pi}{6}$$; $$n = 0$$ gives $$x = \dfrac{5\pi}{6}$$. Both are in the range. $$n = 1$$ gives $$x = \dfrac{17\pi}{6} \approx 8.9 \gt \dfrac{5\pi}{2} \approx 7.85$$, so excluded.

Total number of solutions = 5.

Hence, the correct answer is Option D.