Question

函数y=cos(π/4-2x)的单调递增区间是

Answer 1

y=cos(π/4-2x)=cos(2x-π/4)
2kπ-π<=2x-π/4<=2kπ，则kπ-3π/8<=x<=kπ+π/8
所以，函数的单调递增区间是[kπ-3π/8,kπ+π/8]

追问

2kπ-π<=2x-π/4<=2kπ怎么求出kπ-3π/8<=x<=kπ+π/8

追答

2kπ-π<=2x-π/4<=2kπ
三项都加π/4
2kπ-π+π/4<=2x-π/4+π/4<=2kπ+π/4
2kπ-3π/4<=2x<=2kπ+π/4
kπ-3π/8<=x<=kπ+π/8
你自己是一步也不想算呀

追问

我不是不想自己算 我是真的不会

追答

那你看懂了吗

Answer 2

y=cos(π/4-2x)=cos(2x-π/4)
2kπ-π<=2x-π/4<=2kπ，则kπ-3π/8<=x<=kπ+π/8
所以，函数的单调递增区间是[kπ-3π/8,kπ+π/8]

Answer 3

∵y=cos(x)的单调递增区间是[-π,0]
∴y=cos(π/4-2x)的单调递增区间是
-π≤π/4-2x≤0,
即-5π/4≤-2x≤－π/4
∴π/8≤x≤5π/8

追问

错了亲

追答

π/8≤x≤5π/8

Answer 4

单增区间：-3π/8+kπ<x<π/8 +kπ,k∈z