ĐKXĐ: \(sin2x\ne0\)
Ta có: \(VT=tan^2x+cot^2x=\left(tanx-cotx\right)^2+2\ge2\)
Lại có \(cos^2\left(3x+\frac{\pi}{4}\right)\le1\) ;\(\forall x\Rightarrow VP=1+cos^2\left(3x+\frac{\pi}{4}\right)\le2\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}tanx=cotx\\cos^2\left(3x+\frac{\pi}{4}\right)=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}tanx=tan\left(\frac{\pi}{2}-x\right)\\sin\left(3x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{2}-x+k\pi\\3x+\frac{\pi}{4}=k\pi\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
