f(x)=3sin²x+2√3sinxcosx+cos²x=3(1-cos2x)/2+√3sin2x+(1+cos2x)/2=√3sin2x-cos2x+2=2[sin2x*cos(π/6)-cos2x*sin(π/6)]+2=2sin(2x-π/6)+2(1) 当2x-π/6=2kπ+π/2,k∈Z时,有最大值 4增区间 2kπ-π/...