Câu hỏi 38

\(\begin{array}{l}4{\sin ^4}x + 12{\cos ^2}x - 7 = 0 \Leftrightarrow 4{\sin ^4}x + 12\left( {1 - {{\sin }^2}x} \right) - 7 = 0\\ \Leftrightarrow 4{\sin ^4}x - 12{\sin ^2}x + 5 = 0 \Leftrightarrow \left[ \begin{array}{l}{\sin ^2}x = \dfrac{5}{2}\,\,\left( {loai} \right)\\{\sin ^2}x = \dfrac{1}{2}\,\,\left( {tm} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = \dfrac{{\sqrt 2 }}{2}\\\sin x =  - \dfrac{{\sqrt 2 }}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} + k2\pi \\x = \dfrac{{3\pi }}{4} + k2\pi \\x =  - \dfrac{\pi }{4} + k2\pi \\x = \dfrac{{5\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

Chọn B.