Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{8} +k\pi\\x = \dfrac{3\pi}{8} +k\pi\end{array}\right.\quad (k\in\Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\quad \sin\left(2x + \dfrac{\pi}{3}\right) - \sqrt3\cos\left(2x + \dfrac{\pi}{3}\right)=\sqrt2\\ \Leftrightarrow \dfrac12\sin\left(2x + \dfrac{\pi}{3}\right)-\dfrac{\sqrt3}{2}\cos\left(2x + \dfrac{\pi}{3}\right) = \dfrac{\sqrt2}{2}\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{3}\right)\cos\dfrac{\pi}{3} - \cos\left(2x + \dfrac{\pi}{3}\right)\sin\dfrac{\pi}{3} = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{3} - \dfrac{\pi}{3}\right)=\sin\dfrac{\pi}{4}\\ \Leftrightarrow \sin2x = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{4} +k2\pi\\2x = \dfrac{3\pi}{4} +k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{8} +k\pi\\x = \dfrac{3\pi}{8} +k\pi\end{array}\right.\quad (k\in\Bbb Z)\end{array}$
