Đáp án:
$\left[\begin{array}{l}x=\dfrac{\pi}{12} + k2\pi\\x = \dfrac{19\pi}{36} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\sin\left(2x - \dfrac{\pi}{3}\right)=-\cos\left(x +\dfrac{\pi}{4}\right)$
$\Leftrightarrow \sin\left(2x - \dfrac{\pi}{3}\right)= \cos\left(\dfrac{3\pi}{4}-x\right)$
$\Leftrightarrow \sin\left(2x - \dfrac{\pi}{3}\right)= \sin\left(x -\dfrac{\pi}{4}\right)$
$\Leftrightarrow \left[\begin{array}{l}2x -\dfrac{\pi}{3} =x - \dfrac{\pi}{4} + k2\pi\\2x - \dfrac{\pi}{3} = \dfrac{5\pi}{4} - x + k2\pi\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{12} + k2\pi\\x = \dfrac{19\pi}{36} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)$
