已知函数y=sin(3x+兀/3),求(1)函数的单调区间.

题目

已知函数y=sin(3x+兀/3),求(1)函数的单调区间.
急求

答案

设t=3x+π/3,则y=sin(3x+π/3)=sint的单调递增区间为：2kπ-π/2≤t≤2kπ+π/2,k∈Z也即2kπ-π/2≤3x+π/3≤2kπ+π/2得2kπ/3-5π/18≤x≤2kπ/3+π/18,k∈Z单调递减区间为：2kπ+π/2≤3x+π/3≤2kπ+3π/2得2k...