Đáp án:
a) $\left[\begin{array}{l}x = \dfrac{\pi}{6} +k2\pi\\x = \dfrac{5\pi}{6} +k2\pi\end{array}\right.\quad (k \in \Bbb Z)$
b) $\left[\begin{array}{l}x = -\dfrac{\pi}{12} + k\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right.\quad (k \in \Bbb Z)$
Giải thích các bước giải:
$\begin{array}{l}a)\,\,2\sin^2x - 5\sin x + 2 = 0\\ \Leftrightarrow (\sin x - 2)\left(\sin x - \dfrac{1}{2}\right) =0\\ \Leftrightarrow \left[\begin{array}{l}\sin x = 2\qquad (loại)\\\sin x = \dfrac{1}{2}\quad (nhận)\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} +k2\pi\\x = \dfrac{5\pi}{6} +k2\pi\end{array}\right.\quad (k \in \Bbb Z)\\ b) \,\,\sin2x + \sqrt3\cos2x= 1\\ \Leftrightarrow \dfrac{1}{2}\sin2x + \dfrac{\sqrt3}{2}\cos2x = \dfrac{1}{2}\\ \Leftrightarrow \sin\left(2x + \dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{6}\\ \Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{3} = \dfrac{\pi}{6} + k2\pi\\2x + \dfrac{\pi}{3} = \dfrac{5\pi}{6} +k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = -\dfrac{\pi}{12} + k\pi\\x = \dfrac{\pi}{4} +k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$
