đặt \(t=\tan x+\cot x\)
Thì PT trở thành
\(t^2-2=\dfrac{1}{2}t+1\)
\(\Leftrightarrow2t^2-t-6=0\Leftrightarrow t=2;t=-\dfrac{3}{2}\)
a) TH1 \(t=2\Leftrightarrow\tan x+\cot x=2\Leftrightarrow\tan^2x-2\tan x+1=0\)
\(\Leftrightarrow\tan x=1\Leftrightarrow x=\dfrac{\pi}{4};x=\dfrac{\pi}{4}+\pi\)(vì \(x\in\left(0;2\pi\right)\)
b) TH2:\(t=-\dfrac{3}{2}\Leftrightarrow\tan x+\dfrac{1}{\tan x}=-\dfrac{3}{2}\Leftrightarrow2\tan^2x+3\tan x+1=0\)
\(\Leftrightarrow\tan x=-1;\tan x=-\dfrac{1}{2}\)
+)\(\tan x=-1\Leftrightarrow x=-\dfrac{\pi}{4}+\pi;x=-\dfrac{\pi}{4}+2\pi\)
+) \(\tan x=-\dfrac{1}{2}\Leftrightarrow x=-0,46365+\pi;x=-0,46365+2\pi\)
Vậy trong khoảng đã cho PT có 6 No
