Đáp án:
$\begin{array}{l}
1 + \cos 2x + \sin x = 2{\cos ^2}\dfrac{x}{2}\\
\Rightarrow \cos 2x + \sin x = 2{\cos ^2}\dfrac{x}{2} - 1\\
\Rightarrow \cos 2x + \sin x = \cos \left( {2.\dfrac{x}{2}} \right)\\
\Rightarrow \cos 2x = - \left( {\sin x - \cos x} \right)\\
\Rightarrow {\cos ^2}x - {\sin ^2}x - \left( {\cos x - \sin x} \right) = 0\\
\Rightarrow \left( {\cos x - \sin x} \right)\left( {\cos x + \sin x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\cos x = \sin x\\
\cos x + \sin x = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
\sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = k2\pi \\
x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.
\end{array}$
