Let $$S = \left\{\theta \in (-2\pi, 2\pi) : \cos\theta + 1 = \sqrt{3}\sin\theta\right\}$$. Then the $$\sum_{\theta \in S} \theta$$ is equal to:

$$\cos\theta - \sqrt{3}\sin\theta = -1$$

Divide the equation by $$\sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$$:

$$\frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta = -\frac{1}{2}$$

$$\cos\left(\frac{\pi}{3}\right)\cos\theta - \sin\left(\frac{\pi}{3}\right)\sin\theta = -\frac{1}{2}$$

$$\cos\left(\theta + \frac{\pi}{3}\right) = -\frac{1}{2}$$

$$\theta + \frac{\pi}{3} = 2n\pi \pm \frac{2\pi}{3}$$

$$\theta = 2n\pi + \frac{\pi}{3}$$ or $$\theta = 2n\pi - \pi$$

The set of solutions is $$S = \left\{ -\frac{5\pi}{3}, -\pi, \frac{\pi}{3}, \pi \right\}$$.

$$\sum_{\theta \in S} \theta = -\frac{5\pi}{3} - \pi + \frac{\pi}{3} + \pi$$

$$\sum_{\theta \in S} \theta = -\frac{4\pi}{3}$$