$\begin{array}{l} 1 - \cos \left( {\pi + x} \right) - \sin \left( {\dfrac{{\pi + x}}{2}} \right) = 0\\ \Leftrightarrow 1 - \left( { - \cos x} \right) - \sin \left( {\dfrac{\pi }{2} + \dfrac{x}{2}} \right) = 0\\ \Leftrightarrow 1 + \cos x - \cos \dfrac{x}{2} = 0\\ \Leftrightarrow 2{\cos ^2}\dfrac{x}{2} - \cos \dfrac{x}{2} = 0\\ \Leftrightarrow \cos \dfrac{x}{2}\left( {2\cos \dfrac{x}{2} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos \dfrac{x}{2} = 0\\ 2\cos \dfrac{x}{2} = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \dfrac{x}{2} = \dfrac{\pi }{2} + k\pi \\ \dfrac{x}{2} = \dfrac{\pi }{3} + k2\pi \\ \dfrac{x}{2} = - \dfrac{\pi }{3} + k2\pi \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \pi + k2\pi \\ x = \dfrac{{2\pi }}{3} + k4\pi \\ x = - \dfrac{{2\pi }}{3} + k4\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$
