Câu hỏi 36

\(2\tan x + \cot x - 3 = 0\)   (ĐK: \(x \ne k\pi ;\,\,x \ne \dfrac{\pi }{2} + k\pi \)).

Đặt \(\tan x = t \to 2t + \dfrac{1}{t} - 3 = 0\)

\(\begin{array}{l} \Leftrightarrow 2{t^2} - 3t + 1 = 0 \Leftrightarrow \left[ \begin{array}{l}t = 1\\t = \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\tan x = 1\\\tan x = \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} + k\pi \\x = \arctan \left( {\dfrac{1}{2}} \right) + k\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)