Đáp án:
$\left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x =\dfrac{\pi}{12} + k\pi\\x =\dfrac{5\pi}{12} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$2\sin^22x +\sin2x -1= 0$
$\Leftrightarrow \left(2\sin2x -1\right)(\sin2x +1)=0$
$\Leftrightarrow \left[\begin{array}{l}\sin2x = -1\\\sin2x =\dfrac12\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2x = -\dfrac{\pi}{2} + k2\pi\\2x =\dfrac{\pi}{6} + k2\pi\\2x =\dfrac{5\pi}{6} + k2\p\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{4} + k\pi\\x =\dfrac{\pi}{12} + k\pi\\x =\dfrac{5\pi}{12} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
