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