Phương trình 8cos2xsin2xcos4x = -\(\sqrt 2 \) có nghiệm là:

14/25

\(\left[ \begin{array}{l}x = \frac{{ - \pi }}{{32}} + k\frac{\pi }{4}\\x = \frac{{5\pi }}{{32}} + k\frac{\pi }{4}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).

\(\left[ \begin{array}{l}x = \frac{\pi }{{16}} + k\frac{\pi }{8}\\x = \frac{{3\pi }}{{16}} + k\frac{\pi }{8}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).

\(\left[ \begin{array}{l}x = \frac{\pi }{8} + k\frac{\pi }{8}\\x = \frac{{3\pi }}{8} + k\frac{\pi }{8}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).

\(\left[ \begin{array}{l}x = \frac{\pi }{{32}} + k\frac{\pi }{4}\\x = \frac{{3\pi }}{{32}} + k\frac{\pi }{4}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).

Giải thích

Đáp án đúng là: A

Ta có: 8cos2xsin2xcos4x = -\(\sqrt 2 \) ⇔ 4sin4xcos4x = -\(\sqrt 2 \) ⇔ 2sin8x = -\(\sqrt 2 \)

⇔ sin8x = \( - \frac{{\sqrt 2 }}{2}\)⇔ \(\sin 8x = \sin \left( { - \frac{\pi }{4}} \right) \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{32}} + k\frac{\pi }{4}\\x = \frac{{5\pi }}{{32}} + k\frac{\pi }{4}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\).