(1)f(x)=√3sinxcosx+cos²x
=√3/2sin2x+(cos2x+1)/2
=√3/2sin2x+1/2cos2x+1/2
=sin(2x+π/6)+1/2
f(x)的最小正周期为2π/2=π
2kπ-π/2≤2x+π/6≤2kπ+π/2 (k∈Z)
kπ-π/3≤x≤kπ+π/6
则f(x)的单调递增区间为[kπ-π/3,kπ+π/6] (k∈Z)
(2)2x+π/6=kπ+π/2 (k∈Z)
对称轴为x=kπ/2+π/6
因为0
x0=π/6
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