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

$\begin{array}{l}\sin \mathrm{\theta tan}\mathrm{\theta }+\tan \mathrm{\theta }=\sin 2\mathrm{\theta }\\ \tan \mathrm{\theta }(\sin \mathrm{\theta }+1)=\frac{2\tan \mathrm{\theta }}{1+{\tan }^2\mathrm{\theta }}\\ \tan \mathrm{\theta }=0\Rightarrow \mathrm{\theta }=-\mathrm{\pi },0,\mathrm{\pi }\\ (\sin \mathrm{\theta }+1)=2\cdot {\cos }^2\mathrm{\theta }=2(1+\sin \mathrm{\theta })(1-\sin \mathrm{\theta })\\ \sin \mathrm{\theta }=-1\text{ which is not possible }\\ \sin \mathrm{\theta }=\frac{1}{2}\mathrm{\theta }=\frac{\mathrm{\pi }}{6},\frac{5\mathrm{\pi }}{6}\\ \mathrm{n}\left(\mathrm{s}\right)=5\end{array}$
$\begin{array}{l}\mathrm{T}=\cos 0+\cos 2\mathrm{\pi }+\cos 2\mathrm{\pi }+\cos \frac{\mathrm{\pi }}{3}+\cos \frac{5\mathrm{\pi }}{3}\\ \mathrm{T}=4\\ \mathrm{T}+\mathrm{n}\left(\mathrm{s}\right)=9\end{array}$