问题标题:
函数y=sin(-2x+π/3)的单调递减区间是
问题描述:
函数y=sin(-2x+π/3)的单调递减区间是
房庆辉回答:
∵y=sin(-2x+π/3)
=-sin(2x-π/3)
那么
2kπ-π/2≤2x-π/3≤2kπ+π/2
2kπ-π/2+π/3≤2x≤2kπ+π/2+π/3
2kπ-π/6≤2x≤2kπ+5π/6
kπ-π/12≤x≤kπ+5π/12
∴单调递减区间是[kπ-π/12,kπ+5π/12](k∈z)
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