1、
f(x)=1/2*sin2x-√3(1+cos2x)/2+√3/2+1
=1/2*sin2x-√3/2*cos2x+1
=sin2xcosπ/3-cos2xsinπ/3+1
=sin(2x-π/3)+1
所以T=2π/2=π
sin最小是-1
所以f(x)最小=-1+1=0
此时2x-π/3=2kπ-π/2
所以x∈{x|kπ-π/12,k∈Z}
递增则2kπ-π/2