Câu hỏi 34

\(\begin{array}{l}7\cos x = 4{\cos ^3}x + 4\sin 2x \\ \Leftrightarrow 7\cos x - 4{\cos ^3}x - 8\sin x.\cos x = 0\\ \Leftrightarrow \cos x\left( {7 - 4{{\cos }^2}x - 8\sin x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\7 - 4{\cos ^2}x - 8\sin x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\7 - 4\left( {1 - {{\sin }^2}x} \right) - 8\sin x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\\sin x = \frac{3}{2}(L)\\\sin x = \frac{1}{2}(TM)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\x = \frac{\pi }{6} + k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\left( {k \in Z} \right)\end{array}\)

Chọn B.