f(x)=sin2x+2cos²x-1+1
=sin2x+cos2x+1
=√2(√2/2*sin2x+√2/2cos2x)+1
=√2(sin2xzosπ/4+cos2xsinπ/4)+1
=√2sin(2x+π/4)+1
所以T=2π/2=π
-1≤sin(2x+π/4)≤1
-√2≤√2sin(2x+π/4)≤√2
再加上1
所以值域是[-√2+1,√2+1]