Đáp án:
a) $\left[\begin{array}{l}x = k2\pi\\x= \dfrac{4\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{12}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
a) $2\sin\left(x -\dfrac{\pi}{6}\right) + 1 = 0$
$\to \sin\left(x -\dfrac{\pi}{6}\right) = -\dfrac12$
$\to \sin\left(x -\dfrac{\pi}{6}\right) = \sin\left(-\dfrac{\pi}{6}\right)$
$\to \left[\begin{array}{l}x -\dfrac{\pi}{6} = -\dfrac{\pi}{6} + k2\pi\\x -\dfrac{\pi}{6} = \dfrac{7\pi}{6} + k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x = k2\pi\\x= \dfrac{4\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)$
b) $\sqrt3\cos2x + \sin2x -1 = 0$
$\to \dfrac{\sqrt3}{2}\cos2x +\dfrac12\sin2x =\dfrac12$
$\to \cos\left(2x -\dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{3}$
$\to \left[\begin{array}{l}2x -\dfrac{\pi}{6}=\dfrac{\pi}{3} + k2\pi\\2x -\dfrac{\pi}{6} = -\dfrac{\pi}{3}+ k2\pi\end{array}\right.$
$\to \left[\begin{array}{l}x =\dfrac{\pi}{4} + k\pi\\x = -\dfrac{\pi}{12}+ k\pi\end{array}\right.\quad (k\in\Bbb Z)$
