Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{6} + k\pi\\x =\dfrac{\pi}{2}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\quad 2\sin\left(2x -\dfrac{\pi}{6}\right) - 1= 0$
$\to \sin\left(2x -\dfrac{\pi}{6}\right) =\dfrac12$
$\to \sin\left(2x -\dfrac{\pi}{6}\right) =\sin\dfrac{\pi}{6}$
$\to \left[\begin{array}{l}2x -\dfrac{\pi}{6}=\dfrac{\pi}{6} + k2\pi\\2x -\dfrac{\pi}{6} = \dfrac{5\pi}{6} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}2x =\dfrac{\pi}{3} + k2\pi\\2x =\pi+ k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{6} + k\pi\\x =\dfrac{\pi}{2}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$
