Ta có: \(\text{(1)}\Leftrightarrow 4(1-{\sin}^2 2x)+8\sin 2x-7=0\)
\(\Leftrightarrow 4{\sin}^2 2x-8\sin 2x+3=0\)
\(\Leftrightarrow \left[ \begin{array}{l} \sin 2x = \dfrac{3}{2}>1\text{(loại)}\\\sin 2x=\dfrac{1}{2}\end{array} \right. \)
\(\Leftrightarrow \left[ \begin{array}{l} 2x = \dfrac{\pi}{6}+k2\pi ,k \in \mathbb{Z}\\2x= \pi-({\dfrac{\pi}{6}})+k2\pi ,k \in \mathbb{Z}\end{array} \right. \)
\(\Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi}{12}+k\pi ,k \in \mathbb{Z}\\x= \dfrac{5\pi}{12}+k\pi ,k \in \mathbb{Z}\end{array} \right. \)
Vậy phương trình có nghiệm là \(x = \dfrac{\pi}{12}+k\pi ,k \in \mathbb{Z}\) và \(x= \dfrac{5\pi}{12}+k\pi ,k \in \mathbb{Z}\).
Đáp án: D.
