解由f﹙x﹚=sin ^2x-cos^2x+sin2x-m在［0,π/4］上有零點
即sin ^2x-cos^2x+sin2x-m=0在區(qū)間［0,π/4］有根
即-（cos^2x-sin ^2x）+sin2x-m=0在區(qū)間［0,π/4］有根
即-（cos^2x-sin ^2x）+sin2x=m在區(qū)間［0,π/4］有根
即m=-cos2x+sin2x在區(qū)間［0,π/4］有根
即m=sin2x-cos2x在區(qū)間［0,π/4］有根
即m=√2sin（2x-π/4）在區(qū)間［0,π/4］有根
由x屬于［0,π/4］
即0≤x≤π/4
即0≤2x≤π/2
即-π/4≤2x-π/4≤π/4
故-√2/2≤sin（2x-π/4）≤√2/2
即-1≤√2sin（2x-π/4）≤1
即-1≤m≤1?????????????x?????0,??/4???0??x???/4??0??2x???/2??-??/4??2x-??/4???/4??-??2/2??sin??2x-??/4?????2/2??-1???2sin??2x-??/4????1??-1??m??1??m????[-1,1]?B??