f（x）=2sinx*cosx+2cos^2x-3
=sin(2x)+cos(2x)+1-3
=√2[sin(2x)*√2/2+cos(2x)*√2/2]-2
=√2sin(2x+π/4)-2
所以f(x)的最小正周期T=2π/2=π
当sin(2x+π/4)=1时
f(x)取最大值,为√2-2