Câu hỏi 33

\(2\sin 2x\cos 2x + \sqrt 3 \cos 4x + \sqrt 2  = 0 \Leftrightarrow \sin 4x + \sqrt 3 .\cos 4x =  - \sqrt 2 \)

Chia cả 2 vế cho \(\sqrt {1 + {{\left( {\sqrt 3 } \right)}^2}}  = 2\) , ta có:

\(\begin{array}{l}\frac{1}{2}.\sin 4x + \frac{{\sqrt 3 }}{2}.cos4x =  - \frac{{\sqrt 2 }}{2} \Leftrightarrow \sin \left( {4x + \frac{\pi }{3}} \right) =  - \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l}4x + \frac{\pi }{3} =  - \frac{\pi }{4} + k2\pi \\4x + \frac{\pi }{3} = \pi  - \left( { - \frac{\pi }{4}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{{7\pi }}{{48}} + \frac{{k\pi }}{2}\\x = \frac{{11\pi }}{{48}} + \frac{{k\pi }}{2}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)

KL: \(x \in \left\{ {\frac{{ - 7\pi }}{{48}} + \frac{{k\pi }}{2};\frac{{11\pi }}{{48}} + \frac{{k\pi }}{2},k \in \mathbb{Z}} \right\}\).

Chọn D.