f(x)=(1/2)cos2x-(√3/2)sin2x+1-cos2x =-[(√3/2)sin2x+(1/2)cos2x]+1 =-[sin(π/3)sin2x+cos(π/3)cos2x]+1 =-cos(2x-π/3)+1所以,T=2π/2=π最大值为2单调减区间：π+2kπ