Question

Sol. [Key Concept: Write $cos\theta + 1 = \sqrt{3}sin\theta$ in half-angle form to get $\tan\frac{\theta}{2} = \frac{1}{\sqrt{3}}$; list all $\theta$ in $( - 2\pi,2\pi)$ and add them.]

$${cos\theta + 1 = \sqrt{3}sin\theta }{\frac{1 - \tan^{2}\frac{\theta}{2}}{1 + \tan^{2}\frac{\theta}{2}} + 1 = \sqrt{3}\left( \frac{2tan\frac{\theta}{2}}{1 + \tan^{2}\frac{\theta}{2}} \right) }{2 = 2\sqrt{3}tan\frac{\theta}{2} }{\Rightarrow tan\frac{\theta}{2} = \frac{1}{\sqrt{3}}\ \theta \in ( - 2\pi,2\pi),\frac{\theta}{2} \in ( - \pi,\pi) }{\frac{\theta}{2} = - \frac{5\pi}{6},\frac{\pi}{6} }{\theta = \frac{- 5\pi}{3},\frac{\pi}{3} }$$Sum $= - \frac{- 5\pi}{3} + \frac{\pi}{3} = \frac{- 4\pi}{3}$