Đáp án:
$x \in \left\{ {\dfrac{{7\pi }}{8};\dfrac{{15\pi }}{8};\dfrac{{23\pi }}{8}} \right\}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{\cos ^2}x - \sin 2x = \sqrt 2 + {\cos ^2}\left( {\dfrac{\pi }{2} + x} \right)\\
\Leftrightarrow {\cos ^2}x - \sin 2x = \sqrt 2 + {\left( {\cos \dfrac{\pi }{2}\cos x - \sin \dfrac{\pi }{2}\sin x} \right)^2}\\
\Leftrightarrow {\cos ^2}x - \sin 2x = \sqrt 2 + {\sin ^2}x\\
\Leftrightarrow {\cos ^2}x - {\sin ^2}x - \sin 2x = \sqrt 2 \\
\Leftrightarrow \cos 2x - \sin 2x = \sqrt 2 \\
\Leftrightarrow \dfrac{1}{{\sqrt 2 }}\cos 2x - \dfrac{1}{{\sqrt 2 }}\sin 2x = 1\\
\Leftrightarrow \cos \left( {2x + \dfrac{\pi }{4}} \right) = 1\\
\Leftrightarrow 2x + \dfrac{\pi }{4} = k2\pi \left( {k \in Z} \right)\\
\Leftrightarrow x = - \dfrac{\pi }{8} + k\pi \left( {k \in Z} \right)
\end{array}$
Mà $x \in \left( {0;3\pi } \right) \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{7\pi }}{8}\\
x = \dfrac{{15\pi }}{8}\\
x = \dfrac{{23\pi }}{8}
\end{array} \right.$
Vậy $x \in \left\{ {\dfrac{{7\pi }}{8};\dfrac{{15\pi }}{8};\dfrac{{23\pi }}{8}} \right\}$
