f(x)=√3sin2x+cos2x=2(cosπ/6sin2x+sinπ/6cos2x)=2sin(2x+π/6)单调递增区间为：2kπ-π/2 ≤ 2x+π/6 ≤ 2kπ+π/2解得：kπ-π/3 ≤ x ≤ kπ+π/3单调递减区间为：2kπ+π/2 ≤ 2x+π/6 ≤ 2kπ+3π/2解得：k...对称中心不是x=kπ么？对，2x+π/6=kπ 得：x=kπ/2-π/12所以：原点最近的对称中心的坐标为：(-π/12,0)