TH1: Xét cox = 0 ( có p là nghiệm ko)
TH2: Xét \(\cos x\ne0\). Ta chia cả hai vế \(\cos^2x\)
Pt trở thành \(2\tan^2x-4\tan x+4-1\left(1+\tan^2x\right)=0\)
\(\Leftrightarrow\tan^2x-4\tan x+3=0\)
Đặt \(\tan x=t\). Giải pt nữa là xg ạ
\(2sin^2x-4sinx.cosx+4cos^2x=1\)
\(\Leftrightarrow2\left(sin^2x+cos^2x\right)-4sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow2-2sin2x+cos2x=0\)
\(\Leftrightarrow2sin2x-cos2x=2\)
\(\Leftrightarrow\sqrt{5}\left(\dfrac{2}{\sqrt{5}}sin2x-\dfrac{1}{\sqrt{5}}cos2x\right)=2\)
\(\Leftrightarrow sin\left(2x-arccos\dfrac{2}{\sqrt{5}}\right)=\dfrac{2}{\sqrt{5}}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-arccos\dfrac{2}{\sqrt{5}}=arcsin\dfrac{2}{\sqrt{5}}+k2\pi\\2x-arccos\dfrac{2}{\sqrt{5}}=\pi-arcsin\dfrac{2}{\sqrt{5}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}arccos\dfrac{2}{\sqrt{5}}+\dfrac{1}{2}arcsin\dfrac{2}{\sqrt{5}}+k\pi\\x=\dfrac{\pi}{2}+\dfrac{1}{2}arccos\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}arcsin\dfrac{2}{\sqrt{5}}+k\pi\end{matrix}\right.\)