Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
{\cos ^2}x - \sin 2x = \sqrt 2 + \frac{{1 + \cos \left( {\pi + 2x} \right)}}{2}\\
\to \frac{{1 + \cos 2x}}{2} - \sin 2x = \sqrt 2 + \frac{{1 - \cos 2x}}{2}\\
\to 1 + \cos 2x - 2\sin 2x = 2\sqrt 2 + 1 - \cos 2x\\
\to 2\cos 2x - 2\sin 2x = 2\sqrt 2 \\
\to \cos 2x - \sin 2x = \sqrt 2 \\
\to \frac{1}{{\sqrt 2 }}\cos 2x - \frac{1}{{\sqrt 2 }}\sin 2x = 1\\
\to \sin \frac{\pi }{4}.\cos 2x - \cos \frac{\pi }{4}.\sin 2x = 1\\
\to \sin \left( {\frac{\pi }{4} - 2x} \right) = 1\\
\to \left[ \begin{array}{l}
\frac{\pi }{4} - 2x = \frac{\pi }{2} + k2\pi \\
\frac{\pi }{4} - 2x = - \frac{\pi }{2} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \frac{\pi }{8} + k\pi \\
x = \frac{{3\pi }}{8} + k\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
