f(x)=sinx/2*cosx/2+cos^2(x/2)-2=(1/2)*2sinx/2*cosx/2+[2cos^2(x/2)-1]/2-3/2=(1/2)*sinx+cosx-3/2=√[(1/2)^2+1]*sin(x+arctan2)-3/2=√5/2*sin(1*x+arctan2)-3/2∴函数f(x)的最小正周期是2π/1=2π