ابتدا طرفین وسطین می‌کنیم، داریم: $(1+\cos x)(1-\cos x)=\sin x.\cos \frac{x}{2}\Rightarrow 1-{{\cos }^{2}}x=\sin x.\cos \frac{x}{2}$  $\begin{align}  & \Rightarrow {{\sin }^{2}}x-\sin x.\cos \frac{x}{2}\Rightarrow \sin x(\sin x-\cos \frac{x}{2})=0 \\  & \sin x=0\Rightarrow x=k\pi ,\sin x-\cos \frac{x}{2}=0 \\  & \Rightarrow \sin x=\cos \frac{x}{2} \\  & \Rightarrow \sin x=\sin (\frac{\pi }{2}-\frac{x}{2}) \\ \end{align}$ $\Rightarrow \left\{ \begin{align}  & x=2k\pi +(\frac{\pi }{2}-\frac{x}{2})\Rightarrow x=\frac{4k\pi }{3}+\frac{\pi }{3} \\  & x=2k\pi +\pi -(\frac{\pi }{2}-\frac{x}{2})\Rightarrow x=4k\pi +\pi  \\ \end{align} \right.$