Câu hỏi 17

ĐKXĐ: \(\left\{ \begin{array}{l}4x \ne \dfrac{\pi }{2} + k\pi \\x \ne \dfrac{\pi }{2} + k\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\x \ne \dfrac{\pi }{2} + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

\(\begin{array}{l}\tan 4x.\tan x =  - 1\\ \Leftrightarrow \tan 4x =  - \dfrac{1}{{\tan x}}\\ \Leftrightarrow \tan 4x =  - \cot x\\ \Leftrightarrow \tan 4x =  - \tan \left( {\dfrac{\pi }{2} - x} \right)\\ \Leftrightarrow \tan 4x = \tan \left( {x - \dfrac{\pi }{2}} \right)\\ \Leftrightarrow 4x = x - \dfrac{\pi }{2} + k\pi \\ \Leftrightarrow 3x =  - \dfrac{\pi }{2} + k\pi \\ \Leftrightarrow x =  - \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\,\,\left( {k \in \mathbb{Z}} \right)\,\,\left( {tm} \right)\end{array}\)

Vậy nghiệm của phương trình là \(x =  - \dfrac{\pi }{6} + \dfrac{{k\pi }}{3}\,\,\left( {k \in \mathbb{Z}} \right)\).

Chọn B.