f(x)=(sinx+cosx)²+2cos²x-m
=sin²x+2sinxcosx+cos²x+cos(2x)+1-m
=sin(2x)+cos(2x)+2-m
=√2sin(2x+π/4)+2-m
令f(x)=0
√2sin(2x+π/4)=m-2
x∈R,-1≤sin(2x+π/4)≤1 -√2≤√2sin(2x+π/4)≤√2
要方程有解,-√2≤m-2≤√2
2-√2≤m≤2+√2
m的取值范圍為[2-√2,2+√2]