\(y=1-2\left(2sinx.cosx\right)^2+2sin^42x=1-2sin^22x+2sin^42x\)
\(y=1-2sin^22x\left(1-sin^22x\right)=1-2sin^22x.cos^22x\)
\(=1-\dfrac{1}{2}\left(2sin2x.cos2x\right)^2=1-\dfrac{1}{2}sin^24x\)
Do \(0\le sin^24x\le1\Rightarrow\dfrac{1}{2}\le y\le1\)
\(y_{min}=\dfrac{1}{2}\) khi \(sin^24x=1\)
\(y_{max}=1\) khi \(sin4x=0\)