f(x)=2sin^2(π/4+x)-√3cos2x-1=1-cos(π/2+2x)-√3cos2x-1=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2（1/2sin2x-√3/2cos2x)=2sin(2x-π/3)T=2π/2=πx∈[π/4, π/2],所以(2x-π/3)∈[π/6, 2π/3],那么2sin(...