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PT $\Leftrightarrow \sin 2x = \pm\frac{1}{\sqrt 2}$
Nếu $\sin 2x = \frac{1}{\sqrt 2}=\sin \frac{\pi}{4}\Leftrightarrow \left[ {\begin{matrix} 2x=\frac{\pi}{4}+k2\pi\\ 2x=\frac{3\pi}{4}+k2\pi \end{matrix}} \right.\Leftrightarrow \left[ {\begin{matrix} x=\frac{\pi}{8}+k\pi\\ x=\frac{3\pi}{8}+k\pi \end{matrix}} \right. (k \in \mathbb{Z})$
Nếu $\sin 2x =- \frac{1}{\sqrt 2}=\sin \frac{-\pi}{4}\Leftrightarrow \left[ {\begin{matrix} 2x=\frac{-\pi}{4}+k2\pi\\ 2x=\frac{5\pi}{4}+k2\pi \end{matrix}} \right.\Leftrightarrow \left[ {\begin{matrix} x=\frac{-\pi}{8}+k\pi\\ x=\frac{5\pi}{8}+k\pi \end{matrix}} \right.(k \in \mathbb{Z})$
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