Ta có: \({\rm{(1)}} \Leftrightarrow 4(1 - {\sin ^2}2x) + 8\sin 2x - 7 = 0\)

\(\begin{array}{l}
 \Leftrightarrow 4{\sin ^2}2x - 8\sin 2x + 3 = 0\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{\sin 2x = \frac{3}{2} > 1\,\,{\rm{(l)}}}\\
{\sin 2x = \frac{1}{2}}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{2x = \frac{\pi }{6} + k2\pi ,k \in Z}\\
{2x = \pi  - (\frac{\pi }{6}) + k2\pi ,k \in Z}
\end{array}} \right.\\
 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x = \frac{\pi }{{12}} + k\pi ,k \in Z}\\
{x = \frac{{5\pi }}{{12}} + k\pi ,k \in Z}
\end{array}} \right.
\end{array}\)

Đáp án: D.

-- Mod Toán 11 HỌC247