Let $$S = \{\theta \in (0, 2\pi) : 7\cos^2\theta - 3\sin^2\theta - 2\cos^2(2\theta) = 2\}$$. Then the sum of roots of all the equations $$x^2 - 2(\tan^2\theta + \cot^2\theta)x + 6\sin^2\theta = 0$$, $$\theta \in S$$, is _______

We first solve $$7\cos^2\theta - 3\sin^2\theta - 2\cos^2(2\theta) = 2$$ for $$\theta \in (0, 2\pi)$$.

We write $$\sin^2\theta = 1 - \cos^2\theta$$ and $$\cos^2(2\theta) = (2\cos^2\theta - 1)^2$$. Substituting:

$$7\cos^2\theta - 3(1-\cos^2\theta) - 2(2\cos^2\theta-1)^2 = 2$$

$$10\cos^2\theta - 3 - 2(4\cos^4\theta - 4\cos^2\theta + 1) = 2$$

$$10\cos^2\theta - 3 - 8\cos^4\theta + 8\cos^2\theta - 2 = 2$$

$$-8\cos^4\theta + 18\cos^2\theta - 7 = 0$$

$$8\cos^4\theta - 18\cos^2\theta + 7 = 0$$

Let $$u = \cos^2\theta$$. Then $$8u^2 - 18u + 7 = 0$$. By the quadratic formula:

$$u = \frac{18 \pm \sqrt{324 - 224}}{16} = \frac{18 \pm \sqrt{100}}{16} = \frac{18 \pm 10}{16}$$

So $$u = \frac{28}{16} = \frac{7}{4}$$ or $$u = \frac{8}{16} = \frac{1}{2}$$. Since $$\cos^2\theta \leq 1$$, we discard $$u = 7/4$$ and take $$\cos^2\theta = \frac{1}{2}$$, giving $$\cos\theta = \pm\frac{1}{\sqrt{2}}$$.

The solutions in $$(0, 2\pi)$$ are $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$. So the set $$S$$ contains these four values.

For each $$\theta \in S$$, $$\cos^2\theta = \frac{1}{2}$$ and $$\sin^2\theta = \frac{1}{2}$$. Also $$\tan^2\theta = 1$$ and $$\cot^2\theta = 1$$, so $$\tan^2\theta + \cot^2\theta = 2$$.

The quadratic equation becomes $$x^2 - 2(2)x + 6\cdot\frac{1}{2} = 0$$, i.e., $$x^2 - 4x + 3 = 0$$, which factors as $$(x-1)(x-3) = 0$$. The roots are $$x = 1$$ and $$x = 3$$, with sum $$1 + 3 = 4$$.

Since all four values of $$\theta$$ give the same equation (because they all have the same $$\tan^2\theta + \cot^2\theta$$ and $$\sin^2\theta$$), the sum of roots of ALL equations is $$4 \times 4 = 16$$.

Hence, the correct answer is 16.