b: Đặt a=tan x-cot x
=>\(a^2=\left(\tan x-\cot x\right)^2=\tan^2x+\cot^2x-2\)
=>\(\tan^2x+\cot^2x=a^2+2\)
Ta có: \(\sqrt3\cdot\left(\tan^2x+\cot^2x\right)+2\left(\sqrt3-1\right)\left(\tan x-\cot x\right)-4-2\sqrt3=0\)
=>\(\sqrt3\left(a^2+2\right)+\left(2\sqrt3-2\right)\cdot a-4-2\sqrt3=0\)
=>\(a^2\cdot\sqrt3+\left(2\sqrt3-2\right)\cdot a-4=0\) (1)
\(\Delta=\left(2\sqrt3-2\right)^2-4\cdot\sqrt3\cdot\left(-4\right)=12-8\sqrt3+4+16\sqrt3=16+8\sqrt3\)
\(=\left(2\sqrt3+2\right)^2\) >0
=>(1) có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}a=\frac{-\left(2\sqrt3-2\right)-\sqrt{\left(2\sqrt3+2\right)^2}}{2\cdot\sqrt3}=\frac{-2\sqrt3+2-\left(2\sqrt3+2\right)}{2\sqrt3}=\frac{-4\sqrt3}{2\sqrt3}=-2\\ a=\frac{-\left(2\sqrt3-2\right)+\left(2\sqrt3+2\right)}{2\sqrt3}=\frac{-2\sqrt3+2+2\sqrt3+2}{2\sqrt3}=\frac{4}{2\sqrt3}=\frac{2}{\sqrt3}\end{array}\right.\)
TH1: a=-2
=>\(\tan x-cotx=-2\)
=>\(\tan x-\frac{1}{tanx}=-2\)
=>\(\frac{\tan^2x-1}{\tan x}=-2\)
=>\(\tan^2x-1=-2\cdot tanx\)
=>\(\tan^2x+2\cdot\tan x-1=0\)
=>\(\left(\tan x+1\right)^2=2\)
=>\(\left[\begin{array}{l}tanx+1=\sqrt2\\ \tan x+1=-\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}\tan x=\sqrt2-1\\ tanx=-\sqrt2-1\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\arctan\left(\sqrt2-1\right)+k\pi\\ x=\arctan\left(-\sqrt2-1\right)+k\pi\end{array}\right.\)
TH2: \(a=\frac{2}{\sqrt3}\)
=>\(\tan x-\cot x=\frac{2}{\sqrt3}\)
=>\(\tan x-\frac{1}{\tan x}=\frac{2}{\sqrt3}\)
=>\(\frac{\tan^2x-1}{\tan x}=\frac{2}{\sqrt3}\)
=>\(\sqrt3\cdot\tan^2x-\sqrt3=2\cdot\tan x\)
=>\(\sqrt3\cdot\tan^2x-2\cdot\tan x-\sqrt3=0\)
\(\Delta=\left(-2\right)^2-4\cdot\sqrt3\cdot\left(-\sqrt3\right)=4+4\cdot3=4+12=16>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}\tan x=\frac{2-4}{2\cdot\sqrt3}=\frac{-2}{2\sqrt3}=-\frac{1}{\sqrt3}\\ \tan x=\frac{2+4}{2\sqrt3}=\frac{6}{2\sqrt3}=\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{\pi}{6}+k\pi\\ x=\frac{\pi}{3}+k\pi\end{array}\right.\)
