函数y=2sin(2x+π/4)的单调递增区间怎么求

Question

如题所述

Answer 1

2kπ-π/2<2x+π/4<2kπ+π/2递增 ,整理得kπ-3π/8<x<kπ+π/8
2kπ+π/2<2x+π/4<2kπ+3π/2递减 ， 整理得kπ+π/8<x<kπ+5π/8

Answer 2

-π/2+2kπ≤2x+π/4≤π/2+2kπ
-3π/4≤2x≤π/4
-3π/8≤x≤π/8
单调区间为[-3π/8,π/8]本回答被提问者采纳