Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{12} + \dfrac{\arcsin\dfrac{1}{3}}{3} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{12} - \dfrac{\arcsin\dfrac{1}{3}}{3} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}\sin\left(3x - \dfrac{\pi}{4}\right) = \dfrac{1}{3}\\ \Leftrightarrow \left[\begin{array}{l}3x - \dfrac{\pi}{4} = \arcsin\dfrac{1}{3} + k2\pi\\3x - \dfrac{\pi}{4} = \pi - \arcsin\dfrac{1}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x = \dfrac{\pi}{4} + \arcsin\dfrac{1}{3} + k2\pi\\3x = \dfrac{5\pi}{4} - \arcsin\dfrac{1}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + \dfrac{\arcsin\dfrac{1}{3}}{3} + k\dfrac{2\pi}{3}\\x = \dfrac{5\pi}{12} - \dfrac{\arcsin\dfrac{1}{3}}{3} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
