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