Giải phương trình \({\tan ^2}x + {\cot ^2}x + 2\ta...

A

- Hướng dẫn giải

Lời giải chi tiết:

$\begin{array}{l}\,\,\,\,\,{\tan ^2}x + {\cot ^2}x + 2\tan x + 2cotx = 6\,\,\,\left( {x \ne \dfrac{{k\pi }}{2},\,\,k \in \mathbb{Z}} \right)\\ \Rightarrow {\tan ^2}x + \dfrac{1}{{{{\tan }^2}x}} + 2\left( {\tan x + \dfrac{1}{{\tan x}}} \right) = 6\end{array}$

Đặt $\tan x + \dfrac{1}{{\tan x}} = t \Leftrightarrow {\tan ^2}x + \dfrac{1}{{{{\tan }^2}x}} + 2\tan x.\dfrac{1}{{\tan x}} = {t^2} \Leftrightarrow {\tan ^2}x + \dfrac{1}{{{{\tan }^2}x}} = {t^2} - 2$

Thế vào phương trình ta có:

$\begin{array}{l}{t^2} - 2 + 2t = 6 \Leftrightarrow {t^2} + 2t - 8 = 0 \Leftrightarrow \left[ \begin{array}{l}t = 2\\t =  - 4\end{array} \right.\\ + )\,\,t = 2 \Leftrightarrow \tan x + \dfrac{1}{{\tan x}} = 2 \Leftrightarrow {\tan ^2}x - 2\tan x + 1 = 0\\ \Leftrightarrow \tan x = 1 \Leftrightarrow x = \dfrac{\pi }{4} + k\pi \,\,\,\left( {k \in \mathbb{Z}} \right)\,\,\left( {tm} \right)\\ + )\,\,\,t =  - 4 \Leftrightarrow \tan x + \dfrac{1}{{\tan x}} =  - 4 \Leftrightarrow {\tan ^2}x + 4\tan x + 1 = 0\\ \Leftrightarrow \left[ \begin{array}{l}{\mathop{\rm tanx}\nolimits}  =  - 2 + \sqrt 3  \Leftrightarrow x = \dfrac{{ - \pi }}{{12}} + k\pi \\{\mathop{\rm tanx}\nolimits}  =  - 2 - \sqrt 3  \Leftrightarrow x = \dfrac{{ - 5\pi }}{{12}} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\,\,\left( {tm} \right)\end{array}$

Vậy $S = \left\{ {\dfrac{\pi }{4} + k\pi ,\,\,\dfrac{{ - \pi }}{{12}} + k\pi ;\,\,\dfrac{{ - 5\pi }}{{12}} + k\pi ,\,\,k \in \mathbb{Z}} \right\}$.