$sinx = \dfrac{\sqrt{2}}{2}$
⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{4} + k2\pi\\x=\pi - \dfrac{\pi}{4} + k2\pi = \dfrac{3\pi}{4} + k2\pi\end{array} \right.\)
Do $sinx \ne \dfrac{\sqrt{2}}{2}$
nên $\left \{ {{x\ne\dfrac{\pi}{4} + k2\pi} \atop {x\ne \dfrac{3\pi}{4} + k2\pi}} \right.$