Case I : When cot x > 0, $x \in \left[ {0,{\pi \over 2}} \right] \cup \left[ {\pi ,{{3\pi } \over 2}} \right]$ $\cot x = \cot x + {1 \over {\sin x}} \Rightarrow$ not possible Case II : When cot x < 0, $x \in \left[ {{\pi \over 2},\pi } \right] \cup \left[ {{{3\pi } \over 2},2\pi } \right]$ $- \cot x = \cot x + {1 \over {\sin x}}$ $\Rightarrow {{ - 2\cos x} \over {\sin x}} = {1 \over {\sin x}}$ $\Rightarrow \cos x = {{ - 1} \over 2}$ $\Rightarrow x = {{2\pi } \over 3},{{4\pi } \over 3}$(Rejected) One solution.