[tex]y=2sin2x(sin2x-cos2x)=2sin^{2}2x-2sin2x.cos2x=1-cos4x-sin4x=1-$\sqrt{2}$(cos\alpha .cos4x+sin\alpha .sin4x)=1-$\sqrt{2}$cos(4x-\alpha )[/tex]
với [tex]sin\alpha =cos\alpha[/tex] = 1/[tex]\frac{1}{$\sqrt{2}$}[/tex] $\sqrt{2}$
Mà [tex]-1\leq cos(4x-\alpha )\leq 1 \Rightarrow -$\sqrt{2}$+1\leq y\leq $\sqrt{2}$+1[/tex]