Đáp án:
$\begin{array}{l}
2{\cos ^2}x + {\sin ^2}x - 3\sin x + 2 = 0\\
\Rightarrow 2.\left( {1 - {{\sin }^2}x} \right) + {\sin ^2}x - 3\sin x + 2 = 0\\
\Rightarrow 2 - {\sin ^2}x - 3\sin x + 2 = 0\\
\Rightarrow {\sin ^2}x + 3\sin x - 4 = 0\\
\Rightarrow \left( {\sin x - 1} \right)\left( {\sin x + 4} \right) = 0\\
\Rightarrow \sin x = 1\left( {do: - 1 \le \sin x \le 1} \right)\\
\Rightarrow x = \frac{\pi }{2} + k2\pi
\end{array}$
