Lời giải:
ĐK:..................
PT \(\Leftrightarrow \tan x+\frac{1}{\tan x}+\frac{2}{\tan 2x}(1-2\cos x)=2\)
\(\Leftrightarrow \frac{\tan ^2x+1}{\tan x}+\frac{1-\tan ^2x}{\tan x}(1-2\cos x)=2\)
\(\Leftrightarrow \tan ^2x+1+(1-\tan ^2x)(1-2\cos x)=2\tan x\)
\(\Leftrightarrow (\tan x-1)^2-(\tan x-1)(\tan x+1)(1-2\cos x)=0\)
\(\Leftrightarrow (\tan x-1)[\tan x-1-(\tan x+1)(1-2\cos x)]=0\)
Nếu $\tan x-1=0$ thì $x=k\pi +\frac{\pi}{4}$
Nếu $\tan x-1-(\tan x+1)(1-2\cos x)=0$
$\Leftrightarrow (\tan x+1)\cos x=1$
$\Leftrightarrow (\frac{\sin x}{\cos x}+1)\cos x=1$
$\Leftrightarrow \sin x+\cos x=1$
$\Rightarrow (\sin x+\cos x)^2=1$
$\Leftrightarrow 1+2\sin x\cos x=1$
$\Leftrightarrow \sin x\cos x=0$ (trái điều kiện xác định)
Vậy...............
Lời giải:
ĐK:..................
PT \(\Leftrightarrow \tan x+\frac{1}{\tan x}+\frac{2}{\tan 2x}(1-2\cos x)=2\)
\(\Leftrightarrow \frac{\tan ^2x+1}{\tan x}+\frac{1-\tan ^2x}{\tan x}(1-2\cos x)=2\)
\(\Leftrightarrow \tan ^2x+1+(1-\tan ^2x)(1-2\cos x)=2\tan x\)
\(\Leftrightarrow (\tan x-1)^2-(\tan x-1)(\tan x+1)(1-2\cos x)=0\)
\(\Leftrightarrow (\tan x-1)[\tan x-1-(\tan x+1)(1-2\cos x)]=0\)
Nếu $\tan x-1=0$ thì $x=k\pi +\frac{\pi}{4}$
Nếu $\tan x-1-(\tan x+1)(1-2\cos x)=0$
$\Leftrightarrow (\tan x+1)\cos x=1$
$\Leftrightarrow (\frac{\sin x}{\cos x}+1)\cos x=1$
$\Leftrightarrow \sin x+\cos x=1$
$\Rightarrow (\sin x+\cos x)^2=1$
$\Leftrightarrow 1+2\sin x\cos x=1$
$\Leftrightarrow \sin x\cos x=0$ (trái điều kiện xác định)
Vậy...............
