$\begin{array}{l}
2\sin x{\cos ^2}x = \sin x + 2{\cos ^2}x - 1\\
\Leftrightarrow 2\sin x{\cos ^2}x - 2{\cos ^2}x = \sin x - 1\\
\Leftrightarrow 2{\cos ^2}x\left( {\sin x - 1} \right) - \left( {\sin x - 1} \right) = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {2{{\cos }^2}x - 1} \right) = 0\\
\Leftrightarrow \left( {\sin x - 1} \right).\cos 2x = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\cos 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k2\pi \\
2x = \frac{\pi }{2} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k2\pi \\
x = \frac{\pi }{4} + \frac{{k\pi }}{2}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\alpha = \frac{\pi }{2}\\
\beta = \frac{\pi }{4}
\end{array} \right. \Rightarrow \alpha + \beta = \frac{{3\pi }}{4}
\end{array}$
