Question

数学题啊！！！！函数y=sin(2x-π/3)的单调递增区间

函数y=sin(2x-π/3)的单调递增区间

Answer 1

sinx的减区间是(2kπ+π/2,2kπ+3π/2)
所以2kπ+π/2<2x-π/3<2kπ+3π/2
2kπ+5π/6<2x<2kπ+11π/6
kπ+5π/12<x<kπ+11π/12
所以减区间是(kπ+5π/12，kπ+11π/12)

追问

是增区间QAQ

追答

哦，对不起
那就是2kπ-π/2<2x-π/3<2kπ+π/2
所以是(kπ-π/12,kπ+5π/12)