If $$\theta \in [-2\pi, 2\pi]$$, then the number of solutions of $$2\sqrt{2}\cos^2\theta + (2 - \sqrt{6})\cos\theta - \sqrt{3} = 0$$, is equal to:

We need to solve $$2\sqrt{2}\cos^2\theta + (2 - \sqrt{6})\cos\theta - \sqrt{3} = 0$$ for $$\theta \in [-2\pi, 2\pi]$$.

Let $$u = \cos\theta$$. The equation becomes $$2\sqrt{2}u^2 + (2 - \sqrt{6})u - \sqrt{3} = 0$$.

Using the quadratic formula: $$u = \frac{-(2-\sqrt{6}) \pm \sqrt{(2-\sqrt{6})^2 + 4 \cdot 2\sqrt{2} \cdot \sqrt{3}}}{2 \cdot 2\sqrt{2}}$$.

Computing the discriminant: $$(2-\sqrt{6})^2 + 8\sqrt{6} = 4 - 4\sqrt{6} + 6 + 8\sqrt{6} = 10 + 4\sqrt{6} = (2 + \sqrt{6})^2$$.

So $$u = \frac{(\sqrt{6}-2) \pm (2+\sqrt{6})}{4\sqrt{2}}$$.

Case 1: $$u = \frac{\sqrt{6}-2+2+\sqrt{6}}{4\sqrt{2}} = \frac{2\sqrt{6}}{4\sqrt{2}} = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2}$$.

Case 2: $$u = \frac{\sqrt{6}-2-2-\sqrt{6}}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$$.

For $$\cos\theta = \frac{\sqrt{3}}{2}$$: $$\theta = \pm\frac{\pi}{6}$$. In $$[-2\pi, 2\pi]$$, the solutions are $$\theta = \frac{\pi}{6}, -\frac{\pi}{6}, 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}, -(2\pi - \frac{\pi}{6}) = -\frac{11\pi}{6}$$. That gives 4 solutions.

For $$\cos\theta = -\frac{\sqrt{2}}{2}$$: $$\theta = \pm\frac{3\pi}{4}$$. In $$[-2\pi, 2\pi]$$, the solutions are $$\theta = \frac{3\pi}{4}, -\frac{3\pi}{4}, 2\pi - \frac{3\pi}{4} = \frac{5\pi}{4}, -(2\pi - \frac{3\pi}{4}) = -\frac{5\pi}{4}$$. That gives 4 solutions.

Total number of solutions = $$4 + 4 = 8$$.

Hence, the correct answer is Option C.