Let \(S={\theta \in (-2\pi ,2\pi ):\cos \theta +1=\sqrt{3}\sin \theta }\). Then \(\sum _{\theta \in S}\theta\) is equal …

Given, \(\cos \theta+1=\sqrt{3} \sin \theta\)

\(2 \cos ^2 \frac{\theta}{2}=\sqrt{3} \cdot 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\)

\(2 \cos \frac{\theta}{2}\left(\cos \frac{\theta}{2}-\sqrt{3} \sin \frac{\theta}{2}\right)=0\)

If \(\cos \frac{\theta}{2}=0\)

Now in the interval \((-2 \pi, 2 \pi)\), the possible values are.

\(\theta=-\pi, \pi\)

If \(\cos \frac{\theta}{2}-\sqrt{3} \sin \frac{\theta}{2}=0\)

\(\tan \frac{\theta}{2}=\frac{1}{\sqrt{3}}\)

Now in the interval \((-2 \pi, 2 \pi)\), the possible values are:

\(\theta=-\frac{5 \pi}{3}, \frac{\pi}{3}\)

Therefore,

\(S=\left\{-\pi, \pi,-\frac{5 \pi}{3}, \frac{\pi}{3}\right\}\)

Hence,

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