Đáp án:
$\begin{array}{l}
\sin 2x - 2{\cos ^2}x = \dfrac{1}{2}\\
\Leftrightarrow \sin 2x - 2.\dfrac{{1 + \cos 2x}}{2} = \dfrac{1}{2}\\
\Leftrightarrow \sin 2x - \left( {1 + \cos 2x} \right) = \dfrac{1}{2}\\
\Leftrightarrow \sin 2x - \cos 2x = 1 + \dfrac{1}{2}\\
\Leftrightarrow \sin 2x - \cos 2x = \dfrac{3}{2}\\
\Leftrightarrow \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{3}{2}\\
\Leftrightarrow \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{3\sqrt 2 }}{4}\\
\to \text{Phương trình vô nghiệm}
\end{array}$
