4tan^2(x)-5tan(x)+1=0 có M nghiệm trong khoảng (-2017pi/2;2017pi/2) ?

Đáp án:

Có tất cả 4034 nghiệm.

Giải thích các bước giải:

\(\begin{array}{l} 4{\tan ^2}x – 5\tan x + 1 = 0\,\,\,\,\left( * \right)\\ ĐK:\,\,\,\cos x \ne 0\\ \left( * \right) \Leftrightarrow \left[ \begin{array}{l} \tan x = 1\\ \tan x = \dfrac{1}{4} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{4} + k\pi \\ x = \arctan \dfrac{1}{4} + m\pi \end{array} \right.\,\,\,\left( {k \in Z} \right)\\ x \in \left( { – \dfrac{{2017\pi }}{2};\,\,\dfrac{{2017\pi }}{2}} \right)\\ \Rightarrow \left[ \begin{array}{l} – \dfrac{{2017\pi }}{2} < \,\dfrac{\pi }{4} + k\pi < \dfrac{{2017\pi }}{2}\\ – \dfrac{{2017\pi }}{2} < \,\arctan \dfrac{1}{4} + m\pi < \dfrac{{2017\pi }}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} – 1008,75 < k < 1008,25\\ – 1013,46 < m < 1003,53 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} k \in \left\{ { – 1008; – 1007;…..;\,\,1007;\,\,1008} \right\}\\ m \in \left\{ { – 1013;\,\, – 1012;\,\,….;\,\,1002;\,\,1003} \right\} \end{array} \right.\\ \Rightarrow Có\,\,\,1008.2 + 1 + \left( {1003 + 1013} \right) + 1 = 4034\,\,\,nghiệm. \end{array}\)