已知函数y=sin(3x+兀/3),求(1)函数的单调区间.
已知函数y=sin(3x+兀/3),求(1)函数的单调区间.
急求
急求
数学人气:961 ℃时间:2020-03-25 12:49:10
优质解答
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...
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