Đáp án:
$\left[\begin{array}{l}x = \dfrac{\pi}{12} + k\pi\\x = \dfrac{5\pi}{12} + k\pi\\x = \dfrac{1}{2}\arcsin\dfrac{1}{3} + k\pi\\x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin\dfrac{1}{3} + k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}6\sin^22x - 5\sin2x + 1 = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin2x = \dfrac{1}{2}\\\sin2x = \dfrac{1}{3} \end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = \dfrac{\pi}{6} + k2\pi\\2x = \dfrac{5\pi}{6} + k2\pi\\2x = \arcsin\dfrac{1}{3} + k2\pi\\2x = \pi - \arcsin\dfrac{1}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\pi\\x = \dfrac{5\pi}{12} + k\pi\\x = \dfrac{1}{2}\arcsin\dfrac{1}{3} + k\pi\\x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin\dfrac{1}{3} + k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
