\(a\left( {{{\tan }^2}x + {{\cot }^2}x} \right) + b\left( {\tan x \pm \cot x} \right) + c = 0\).

Đặt \(t = \tan x \pm \cot x = \frac{{\sin x}}{{\cos x}} \pm \frac{{\cos x}}{{\sin x}} = \left\langle {\begin{array}{*{20}{c}}{\frac{1}{{\sin x\cos x}} = \frac{2}{{\sin 2x}}}\\{\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}} =  - \frac{{\cos 2x}}{{\sin 2x}}}\end{array}} \right.\)

Lại có \({t^2} = {\tan ^2}x + {\cot ^2}x \pm 2 \Rightarrow {\tan ^2}x + {\cot ^2}x = {t^2} \mp 2\).

Thay vào phương trình ẩn t, tìm t rồi suy ra x.

Giải phương trình:

a) \((\tan x + 7)\tan x + (\cot x + 7)\cot x + 14 = 0\);

b) \(\sqrt 3 ({\tan ^2}x + {\cot ^2}x) + 2(\sqrt 3  - 1)(\tan x - \cot x) - 4 - 2\sqrt 3  = 0\).

Giải:

a) Điều kiện: \(\left\{ \begin{array}{l}\sin x \ne 0\\\cos x \ne 0\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow x \ne \frac{{k\pi }}{2}\) \(\left( {k \in \mathbb{Z}} \right)\).

\((\tan x + 7)\tan x + (\cot x + 7)\cot x + 14 = 0\)

\( \Leftrightarrow ({\tan ^2}x + {\cot ^2}x) + 7(\tan x + \cot x) + 14 = 0\).

Đặt \(\tan x + \cot x = t\), \(\left| t \right| \ge 2\), suy ra \({\tan ^2}x + {\cot ^2}x = {t^2} - 2\).

Khi đó, phương trình có dạng \({t^2} - 2 + 7t + 14 = 0\)

\( \Leftrightarrow {t^2} + 7t + 12 = 0 \Leftrightarrow \left[ \begin{array}{l}t =  - 3\\t =  - 4\end{array} \right.\)

+ Với t = -3, ta có:

\(\tan x + \cot x =  - 3 \Leftrightarrow \tan x + \frac{1}{{\tan x}} =  - 3 \Leftrightarrow {\tan ^2}x + 3\tan x + 1 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\tan x = \frac{{ - 3 - \sqrt 5 }}{2} = \tan \alpha \\\tan x = \frac{{ - 3 + \sqrt 5 }}{2} = \tan \beta \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \alpha  + k\pi \\x = \beta  + k\pi \end{array} \right.\) \(\left( {k \in \mathbb{Z}} \right)\).

+ Với t = -4, ta có:

\(\tan x + \cot x =  - 4 \Leftrightarrow \frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\sin x}} =  - 4 \Leftrightarrow \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x\cos x}} =  - 4\)

\( \Leftrightarrow \sin 2x =  - \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l}2x =  - \frac{\pi }{6} + k2\pi \\2x = \frac{{7\pi }}{6} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{\pi }{{12}} + k\pi \\x = \frac{{7\pi }}{{12}} + k\pi \end{array} \right.\) \(\left( {k \in \mathbb{Z}} \right)\).

b) Điều kiện: \(\left\{ \begin{array}{l}\sin x \ne 0\\\cos x \ne 0\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow x \ne \frac{{k\pi }}{2}\) \(\left( {k \in \mathbb{Z}} \right)\).

Đặt \(\tan x - \cot x = t\), suy ra \({\tan ^2}x + {\cot ^2}x = {t^2} + 2\).

Khi đó, phương trình có dạng \(\sqrt 3 ({t^2} + 2) + 2(\sqrt 3  - 1)t - 4 - 2\sqrt 3  = 0\)

\( \Leftrightarrow \sqrt 3 {t^2} + 2(\sqrt 3  - 1)t - 4 = 0 \Leftrightarrow \left[ \begin{array}{l}t =  - 2\\t = \frac{2}{{\sqrt 3 }}\end{array} \right.\)

+ Với \(t = \frac{2}{{\sqrt 3 }}\), ta có:

\(\tan x - \cot x = \frac{2}{{\sqrt 3 }} \Leftrightarrow \frac{{\sin x}}{{\cos x}} - \frac{{\cos x}}{{\sin x}} = \frac{2}{{\sqrt 3 }}\)

\( \Leftrightarrow \frac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x\cos x}} = \frac{2}{{\sqrt 3 }} \Leftrightarrow \cot 2x =  - \frac{1}{{\sqrt 3 }}\)

\( \Leftrightarrow 2x =  - \frac{\pi }{3} + k\pi  \Leftrightarrow x =  - \frac{\pi }{6} + \frac{{k\pi }}{2}\) \(\left( {k \in \mathbb{Z}} \right)\).

+ Với \(t =  - 2\), ta có:

\(\tan x - \cot x =  - 2 \Leftrightarrow \tan x - \frac{1}{{\tan x}} =  - 2\)

\( \Leftrightarrow {\tan ^2}x + 2\tan x - 1 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\tan x =  - 1 - \sqrt 2  = \tan \alpha \\\tan x =  - 1 + \sqrt 2  = \tan \beta \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \alpha  + k\pi \\x = \beta  + k\pi \end{array} \right.\) \(\left( {k \in \mathbb{Z}} \right)\).