Đặt \(\left\{{}\begin{matrix}cosx=a\\\sqrt{2-cos^2x}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=2\\a+b+ab=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=2\\2\left(a+b\right)+2ab=6\end{matrix}\right.\)
\(\Rightarrow\left(a+b\right)^2+2\left(a+b\right)-8=0\Rightarrow\left[{}\begin{matrix}a+b=2\\a+b=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cosx+\sqrt{2-cos^2x}=2\)
Mà \(cosx+\sqrt{2-cos^2x}\le\sqrt{2\left(cos^2x+2-cos^2x\right)}=2\)
Dấu "=" xảy ra khi và chỉ khi:
\(cosx=1\Leftrightarrow x=k2\pi\)
