f(x)=2√3sinxcosx+2cos^2x-1=√3sin(2x)+cos(2x)=2[sin(2x)(√3/2)+cos(2x)(1/2)]=2[sin(2x)cos(π/6)+cos(2x)sin(π/6)]=2sin(2x+π/6)=2sin[2(x+π/12)]最小正周期：π,当x=π/6时,有最大值2,当x=π/2时,有最小值...