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

We need to solve: $$3\cos^4\theta - 5\cos^2\theta - 2\sin^6\theta + 2 = 0$$

Let $$c = \cos^2\theta$$. Then $$\sin^2\theta = 1 - c$$.

$$\sin^6\theta = (1-c)^3 = 1 - 3c + 3c^2 - c^3$$

Substituting:

$$3c^2 - 5c - 2(1 - 3c + 3c^2 - c^3) + 2 = 0$$

$$3c^2 - 5c - 2 + 6c - 6c^2 + 2c^3 + 2 = 0$$

$$2c^3 - 3c^2 + c = 0$$

$$c(2c^2 - 3c + 1) = 0$$

$$c(2c - 1)(c - 1) = 0$$

So $$\cos^2\theta = 0$$, $$\cos^2\theta = \frac{1}{2}$$, or $$\cos^2\theta = 1$$.

Case 1: $$\cos^2\theta = 0$$ gives $$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$ — 2 solutions

Case 2: $$\cos^2\theta = \frac{1}{2}$$ gives $$\cos\theta = \pm\frac{1}{\sqrt{2}}$$ — $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$ — 4 solutions

Case 3: $$\cos^2\theta = 1$$ gives $$\theta = 0, \pi, 2\pi$$ — 3 solutions

Total number of elements in set S = 2 + 4 + 3 = 9.