Let $\theta , \varnothing \in [0,2\pi ]$ be such that $2\cos \theta \left(1-\sin \varnothing \right) = {\sin }^2 \theta \left(\tan \frac{\theta }{2}+ cot\frac{\theta }{2}\right) \cos \varnothing -1$,
$\tan (2\pi - \theta ) >0 and -1< \sin \theta <-\frac{\sqrt{3}}{2}$ . then $\varnothing$ can’t satisfy 

$$\begin{gathered} \tan (2\pi -\theta )>0\Rightarrow -\tan \theta >0\Rightarrow \tan \theta <0 \\ \begin{array}{l}\Rightarrow \theta \in IIorIVquadrant\\ -1<\sin \theta <-\frac{\sqrt{3}}{2}\Rightarrow \frac{4\pi }{3}<\theta <\frac{5\pi }{3}-\left\{\frac{3\pi }{2}\right\}\end{array} \\ \begin{array}{l}\Rightarrow \frac{3\pi }{2}<\theta <\frac{5\pi }{3}\mathrm{.......}\left(i\right)\\ 2\cos \theta (1-\sin \varphi )={\sin }^2\theta \left(\tan \frac{\theta }{2}+\cot \frac{\theta }{2}\right)\cos \varphi -1\end{array} \\ \Rightarrow 2\cos \theta -2\cos \theta .\sin \varphi ={\sin }^2\theta \left(\frac{{\sin }^2\frac{\theta }{2}+{\cos }^2\theta /2}{\sin \theta /2.\cos \theta /2}\right)\cos \varphi -1 \\ \begin{array}{l}\Rightarrow 2\cos \theta -2\cos \theta .\sin \varphi =2{\sin }^2\theta \left(\frac{1}{\sin \theta }\right).\cos \varphi -1\\ \Rightarrow 2\cos \theta +1=2\sin (\theta +\varphi )\end{array} \\ \begin{array}{l}\frac{3\pi }{2}<\theta <\frac{5\pi }{3}\Rightarrow 2\cos \theta +1\in (1,2)\\ 1<2\sin (\theta +\varphi )<2\Rightarrow \frac{1}{2}<\sin (\theta +\varphi )<1\end{array} \\ \begin{array}{l}\Rightarrow \frac{\pi }{6}<\theta +\varphi <\frac{5\pi }{6}or\frac{13\pi }{6}<\theta +\varphi <\frac{17\pi }{6}-\left\{\frac{\pi }{2}\right\}\\ \Rightarrow \frac{\pi }{6}-\theta <\varphi <\frac{5\pi }{6}-\theta or\frac{13\pi }{6}-\theta <\varphi <\frac{17\pi }{6}-\theta \\ \Rightarrow \varphi \in (-\frac{3\pi }{2},-\frac{2\pi }{3})or(\frac{2\pi }{3},\frac{7\pi }{6})\end{array} \end{gathered}$$