(1) f(x)=(1-cos2x)/2+√3/2*sin2x+1+cos2x=√3/2*sin2x+1/2*cos2x+3/2=sin(2x+π/6)+3/2所以f(x)的最小正周期为π在(kπ-π/3,kπ+π/6)上单调递增，在(kπ+π/6,kπ+2π/3)上单调递减(2) f(x)可以由函数y=sin2x的图像向左平移π/12，向上平移3/2得到
