设3x+pai/4=t,sint的单调增区间是[-2/pai+2kpai, 2/pai+2kpai], 求解出x即可
3x+π/4在[2kπ-π/2,2kπ+π/2]递增
x在[2kπ/3-π/4,2kπ/3+π/12]递增
解：
-π/2+2kπ≤3x+π/4≤π/2+2kπ，k∈z
-3π/4+2kπ≤3x≤π/4+2kπ，k∈z
-π/4+2kπ/3≤x≤π/12+2kπ/3，k∈z
∴f（x）的递增区间为【-π/4+2kπ/3,π/12+2kπ/3】.k∈z
答题不易，且回且珍惜
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