The number of elements in the set $$S = \{\theta \in [-4\pi, 4\pi] : 3\cos^2 2\theta + 6\cos 2\theta - 10\cos^2\theta + 5 = 0\}$$ is ______.

We need to find the number of elements in $$S = \{\theta \in [-4\pi, 4\pi] : 3\cos^2 2\theta + 6\cos 2\theta - 10\cos^2\theta + 5 = 0\}$$. Using $$\cos 2\theta = 2\cos^2\theta - 1$$ gives $$\cos^2\theta = \frac{\cos 2\theta + 1}{2}$$, and substituting yields $$3\cos^2 2\theta + 6\cos 2\theta - 10 \cdot \frac{\cos 2\theta + 1}{2} + 5 = 0$$ which simplifies to $$3\cos^2 2\theta + \cos 2\theta = 0$$ and thus $$\cos 2\theta(3\cos 2\theta + 1) = 0$$.

From $$\cos 2\theta = 0$$ we obtain $$2\theta = \frac{\pi}{2} + n\pi$$ and hence $$\theta = \frac{\pi}{4} + \frac{n\pi}{2}$$ for integer $$n$$. Enforcing $$-4\pi \leq \theta \leq 4\pi$$ gives $$-\frac{17}{2} \leq n \leq \frac{15}{2}$$, so $$n = -8, -7, \ldots, 7$$, yielding 16 solutions.

Solving $$\cos 2\theta = -\frac{1}{3}$$ leads to $$2\theta = \pm \cos^{-1}\left(-\frac{1}{3}\right) + 2k\pi$$ and hence $$\theta = \pm \frac{1}{2}\cos^{-1}\left(-\frac{1}{3}\right) + k\pi$$. Let $$\phi = \frac{1}{2}\cos^{-1}\left(-\frac{1}{3}\right)$$, so the two solution families are $$\theta = \phi + k\pi$$ and $$\theta = -\phi + k\pi$$. Each has period $$\pi$$, giving 8 solutions per family on $$[-4\pi,4\pi]$$, for a total of 16. Since $$\cos^{-1}(-1/3)$$ is not a rational multiple of $$\pi$$, there is no overlap with the previous solutions.

Therefore, $$|S| = 16 + 16 = 32$$. The answer is $$\boxed{32}$$.