1）y取最大值时,3x+π/6=2kπ+π/2(k∈Z) 即x=2kπ/3+π/9(k∈Z)；
y取最小值时,3x+π/6＝2kπ-π/2(k∈Z) 即x=2kπ/3-2π/9(k∈Z).
2)令 2kπ－π/2≤3x+π/6≤2kπ+π/2,得：2kπ/3-2π/3≤x≤2kπ+π/9(k∈Z)；
令 2kπ+π/2≤3x+π/6≤2kπ+π+π/2,得：2kπ/3+π/9≤x≤2kπ/3+4π/9(K∈Z).
故此函数的单调递增区间为：[2kπ/3-2π/3,2kπ+π/9](K∈Z)；
单调递减区间为：[2kπ/3+π/9,2kπ/3+4π/9](K∈Z).