高中数学三角函数求单调区间问题

2025-04-03 14:41:52
推荐回答(1个)
回答1:

sin(2x-π/3)的单调区间是-π/2+2kπ≤2x-π/3≤π/2+2kπ
然后变成-π/2+2kπ+π/3≤2x≤π/2+2kπ +π/3
-π/4+kπ+π/6≤x≤π/4+kπ +π/6即-π/12+kπ≤x≤5π/12+kπ
sin(x/2-π/3)的单调区间是-π/2+2kπ≤x/2-π/3≤π/2+2kπ
-π/2+2kπ+π/3≤x/2≤π/2+2kπ +π/3
-π+4kπ+2π/3≤x≤π+4kπ +2π/3
其中的4kπ不可以改变