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

We need to solve $$\cos\left(x + \frac{\pi}{3}\right)\cos\left(\frac{\pi}{3} - x\right) = \frac{1}{4}\cos^2 2x$$ in $$[-3\pi, 3\pi]$$. Simplify the left-hand side using the product-to-sum formula $$\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)],$$ where $$A = x + \frac{\pi}{3}$$ and $$B = \frac{\pi}{3} - x$$, so that $$A - B = 2x,\quad A + B = \frac{2\pi}{3}$$ and hence $$\text{LHS} = \frac{1}{2}\left[\cos 2x + \cos\frac{2\pi}{3}\right] = \frac{1}{2}\left[\cos 2x - \frac{1}{2}\right].$$

The equation becomes $$\frac{1}{2}\left[\cos 2x - \frac{1}{2}\right] = \frac{1}{4}\cos^2 2x$$ and multiplying both sides by 4 gives $$2\cos 2x - 1 = \cos^2 2x.$$ Rearranging leads to $$\cos^2 2x - 2\cos 2x + 1 = 0,$$ i.e. $$(\cos 2x - 1)^2 = 0,$$ so $$\cos 2x = 1,$$ which in turn yields $$2x = 2n\pi$$ and hence $$x = n\pi$$ for integer $$n$$.

Within $$[-3\pi, 3\pi]$$ this gives $$x = -3\pi, -2\pi, -\pi, 0, \pi, 2\pi, 3\pi,$$ a total of 7 solutions. Therefore, the answer is Option D: 7.