Đáp án đúng: B
Giải chi tiết:ĐKXĐ: \(\left\{ \begin{array}{l}\cos x \ne 0\\\cos 2x \ne 0\\\cos 4x \ne 0\end{array} \right.\)
\(\begin{array}{l}\tan 4x - \tan 2x - 4\tan \,x = 4\tan 4x.\tan 2x.\tan \,x\\ \Leftrightarrow \tan 4x - \tan 2x = 4\tan 4x.\tan 2x.\tan \,x + 4\tan \,x\\ \Leftrightarrow \tan 4x - \tan 2x = 4\tan \,x\left( {\tan 4x.\tan 2x + 1} \right)\\ \Leftrightarrow \frac{{\tan 4x - \tan 2x}}{{\tan 4x.\tan 2x + 1}} = 4\tan \,x \Leftrightarrow \tan 2x = 4\tan x\\ \Leftrightarrow \frac{{2\tan \,x}}{{1 - {{\tan }^2}x}} = 4\tan x \Leftrightarrow 2\tan \,x\left( {1 - 2 + 2{{\tan }^2}x} \right) = 0\\ \Leftrightarrow 2\tan \,x\left( {2{{\tan }^2}x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\tan \,x = 0\\{\tan ^2}x = \frac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\tan \,x = 0\\\frac{{1 - \cos 2x}}{{1 + \cos 2x}} = \frac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\tan \,x = 0\\\cos 2x = \frac{1}{3}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \pm \frac{1}{2}\arccos \frac{1}{3} + k\pi \end{array} \right.,k \in Z\end{array}\)
+) \(x = k\pi ,k \in Z,\,\,x \in \left[ { - \pi ;\pi } \right] \Rightarrow x \in \left\{ { - \pi ;0;\pi } \right\}\)
+) \(x = \frac{1}{2}\arccos \frac{1}{3} + k\pi ,k \in Z,\,\,x \in \left[ { - \pi ;\pi } \right] \Rightarrow x \in \left\{ {\frac{1}{2}\arccos \frac{1}{3};\frac{1}{2}\arccos \frac{1}{3} - \pi } \right\}\)
+) \(x = - \frac{1}{2}\arccos \frac{1}{3} + k\pi ,k \in Z,\,\,x \in \left[ { - \pi ;\pi } \right] \Rightarrow x \in \left\{ { - \frac{1}{2}\arccos \frac{1}{3}; - \frac{1}{2}\arccos \frac{1}{3} + \pi } \right\}\)
Vậy, phương trình có tất cả 7 nghiệm thuộc đoạn \(\left[ { - \pi ;\pi } \right]\).
Chọn: B
