Đáp án:
$\left[\begin{array}{l}x =\dfrac{\pi}{12} + k\pi\\x = \dfrac{3\pi}{4} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sqrt2\sin\left(2x - \dfrac{\pi}{3}\right) - \cos\dfrac{3\pi}{4}=0$
$\Leftrightarrow \sqrt2\sin\left(2x - \dfrac{\pi}{3}\right) =-\dfrac{\sqrt2}{2}$
$\Leftrightarrow \sin\left(2x - \dfrac{\pi}{3}\right) =-\dfrac{1}{2}$
$\Leftrightarrow \left[\begin{array}{l}2x - \dfrac{\pi}{3} = -\dfrac{\pi}{6} + k2\pi\\2x -\dfrac{\pi}{3} = \dfrac{7\pi}{6} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}2x =\dfrac{\pi}{6} + k2\pi\\2x = \dfrac{3\pi}{2} + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x =\dfrac{\pi}{12} + k\pi\\x = \dfrac{3\pi}{4} + k\pi\end{array}\right.\quad (k\in\Bbb Z)$
