If $$\sqrt{3}(\cos^2 x) = (\sqrt{3} - 1)\cos x + 1$$, then number of solutions of the given equation when $$x \in \left[0, \frac{\pi}{2}\right]$$ is ______.

We need to solve $$\sqrt{3}\cos^2 x = (\sqrt{3} - 1)\cos x + 1$$ for $$x \in \left[0, \frac{\pi}{2}\right]$$.

Rearranging: $$\sqrt{3}\cos^2 x - (\sqrt{3} - 1)\cos x - 1 = 0$$.

Let $$t = \cos x$$. Then $$\sqrt{3}t^2 - (\sqrt{3} - 1)t - 1 = 0$$.

Using the quadratic formula: $$t = \frac{(\sqrt{3}-1) \pm \sqrt{(\sqrt{3}-1)^2 + 4\sqrt{3}}}{2\sqrt{3}}$$.

The discriminant is $$(\sqrt{3}-1)^2 + 4\sqrt{3} = 3 - 2\sqrt{3} + 1 + 4\sqrt{3} = 4 + 2\sqrt{3} = (\sqrt{3}+1)^2$$.

So $$t = \frac{(\sqrt{3}-1) \pm (\sqrt{3}+1)}{2\sqrt{3}}$$.

Taking the positive sign: $$t = \frac{2\sqrt{3}}{2\sqrt{3}} = 1$$, so $$\cos x = 1$$, giving $$x = 0$$.

Taking the negative sign: $$t = \frac{(\sqrt{3}-1) - (\sqrt{3}+1)}{2\sqrt{3}} = \frac{-2}{2\sqrt{3}} = \frac{-1}{\sqrt{3}}$$.

Since $$x \in [0, \frac{\pi}{2}]$$, we need $$\cos x \geq 0$$, so $$\cos x = -\frac{1}{\sqrt{3}}$$ gives no valid solution in this interval.

Therefore the only solution is $$x = 0$$, and the number of solutions is $$1$$.