Let $$S = \left\{\theta \in [-\pi, \pi] - \left\{\pm \frac{\pi}{2}\right\} : \sin\theta \tan\theta + \tan\theta = \sin 2\theta\right\}$$. If $$T = \sum_{\theta \in S} \cos 2\theta$$, then $$T + n(S)$$ is equal to

We need to solve: $$\sin\theta\tan\theta + \tan\theta = \sin 2\theta$$ for $$\theta \in [-\pi, \pi] \setminus \{\pm\frac{\pi}{2}\}$$.

Rewriting the equation:

$$ \tan\theta(\sin\theta + 1) = 2\sin\theta\cos\theta $$

$$ \frac{\sin\theta}{\cos\theta}(\sin\theta + 1) = 2\sin\theta\cos\theta $$

Case 1: $$\sin\theta = 0$$

LHS = 0, RHS = 0. This gives $$\theta = -\pi, 0, \pi$$.

Note: $$\theta = -\pi$$ and $$\theta = \pi$$ represent the same point. We include both if both are in the domain. Since the domain is $$[-\pi, \pi]$$, both $$-\pi$$ and $$\pi$$ are included.

So $$\theta \in \{-\pi, 0, \pi\}$$.

Case 2: $$\sin\theta \neq 0$$. Divide both sides by $$\sin\theta$$:

$$ \frac{\sin\theta + 1}{\cos\theta} = 2\cos\theta $$

$$ \sin\theta + 1 = 2\cos^2\theta = 2(1 - \sin^2\theta) = 2 - 2\sin^2\theta $$

$$ 2\sin^2\theta + \sin\theta - 1 = 0 $$

$$ (2\sin\theta - 1)(\sin\theta + 1) = 0 $$

$$ \sin\theta = \frac{1}{2} \quad \text{or} \quad \sin\theta = -1 $$

$$\sin\theta = -1$$ gives $$\theta = -\frac{\pi}{2}$$, which is excluded from the domain.

$$\sin\theta = \frac{1}{2}$$ gives $$\theta = \frac{\pi}{6}$$ or $$\theta = \frac{5\pi}{6}$$.

We must verify $$\cos\theta \neq 0$$ for these values (already satisfied).

So $$S = \left\{-\pi, 0, \pi, \frac{\pi}{6}, \frac{5\pi}{6}\right\}$$

$$n(S) = 5$$

Now compute $$T = \sum_{\theta \in S} \cos 2\theta$$:

$$\cos 2(-\pi) = \cos(-2\pi) = 1$$

$$\cos 2(0) = \cos 0 = 1$$

$$\cos 2(\pi) = \cos 2\pi = 1$$

$$\cos 2\left(\frac{\pi}{6}\right) = \cos\frac{\pi}{3} = \frac{1}{2}$$

$$\cos 2\left(\frac{5\pi}{6}\right) = \cos\frac{5\pi}{3} = \frac{1}{2}$$

$$ T = 1 + 1 + 1 + \frac{1}{2} + \frac{1}{2} = 4 $$

$$ T + n(S) = 4 + 5 = 9 $$

The answer is Option D: 9.