已知函数2√3sinxcosx+cos2x（x∈R）问（1）函数的最小正周期是多少,并求出其一条对称轴方程

f(x)=√3sin2x+cos2x
=2×(√3/2×sin2x+1/2×cos2x)
=2sin(2x+π/6)
(1)最小正周期T=2π/2=π
令2x+π/6=π/2,那么x=π/6,于是有1条对称轴为x=π/6
(2)令2kπ-π/2≤2x+π/6≤2kπ+π/2
那么kπ-π/3≤2x≤kπ+π/6,即单调递增区间为[kπ-π/3,kπ+π/6] (k∈Z)
(3)y=2sinx的图像向左平移π/6个单位————→y=2sin(x+π/6),然后纵坐标不变,横坐标变为原来的1/2————→y=2sin(2x+π/6)