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

Đặt $t=\tan x\pm \cot x=\frac{\sin x}{\cos x}\pm \frac{\cos x}{\sin x}=⟨\begin{array}{c}\frac{1}{\sin x\cos x}=\frac{2}{\sin 2x}\\ \frac{{\sin }^{2}x-{\cos }^{2}x}{\sin x\cos x}=-\frac{\cos 2x}{\sin 2x}\end{array}$

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: $\{\begin{array}{c}\sin x\ne 0\\ \cos x\ne 0\end{array}\iff \sin 2x\ne 0\iff x\ne \frac{k\pi }{2}$ $\left(k\in Z\right)$.

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

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

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 $t^2-2+7t+14=0$

$\iff t^2+7t+12=0\iff [\begin{array}{c}t=-3\\ t=-4\end{array}$

+ Với t = -3, ta có:

$\tan x+\cot x=-3\iff \tan x+\frac{1}{\tan x}=-3\iff {\tan }^{2}x+3\tan x+1=0$

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

+ Với t = -4, ta có:

$\tan x+\cot x=-4\iff \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=-4\iff \frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}=-4$

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

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

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

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

$\iff \sqrt{3}{t}^2+2(\sqrt{3}-1)t-4=0\iff [\begin{array}{c}t=-2\\ t=\frac{2}{\sqrt{3}}\end{array}$

+ Với $t=\frac{2}{\sqrt{3}}$, ta có:

$\tan x-\cot x=\frac{2}{\sqrt{3}}\iff \frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}=\frac{2}{\sqrt{3}}$

$\iff \frac{{\sin }^{2}x-{\cos }^{2}x}{\sin x\cos x}=\frac{2}{\sqrt{3}}\iff \cot 2x=-\frac{1}{\sqrt{3}}$

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

+ Với $t=-2$, ta có:

$\tan x-\cot x=-2\iff \tan x-\frac{1}{\tan x}=-2$

$\iff {\tan }^{2}x+2\tan x-1=0$

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