答：f(x)=sin2x-2√3(sinx)^2-√3+1=sin2x+√3(1-2(sinx)^2]-2√3+1=sin2x+√3cos2x-2√3+1=2*[(1/2)sin2x+(√3/2)cos2x]-2√3+1=2sin(2x+π/3)-2√3+1最小正周期T=2π/2=π单调递增区间满足：2kπ-π/2...