Đáp án:$\left[ \begin{array}{l}
x = k\pi \\
x = \frac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in Z} \right)$
Giải thích các bước giải:
$\begin{array}{l}
2{\cos ^2}x + \sin 2x - 2 = 0\\
\Rightarrow 2{\cos ^2}x - 1 + \sin 2x - 1 = 0\\
\Rightarrow \cos 2x + \sin 2x = 1\\
\Rightarrow \sqrt 2 \sin \left( {2x + \frac{\pi }{4}} \right) = 1\\
\Rightarrow \sin \left( {2x + \frac{\pi }{4}} \right) = \frac{1}{{\sqrt 2 }}\\
\Rightarrow \left[ \begin{array}{l}
2x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\
2x + \frac{\pi }{4} = \pi - \frac{\pi }{4} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \frac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}$
