求函数y=√2sin(2x-π⼀4)的单调递减区间

2025-02-23 22:43:44
推荐回答(3个)
回答1:

y=2sin(2x+π/4)+1 y=sinx单调递减区间是 【2kπ-π/2,2kπ+π/2】所以2kπ-π/2<=2x+π/4<=2kπ+π/2 kπ-3π/8<=x<=kπ+π/8

回答2:

y=-2sin(2x-π/4)
y递减则sin(2x-π/4)递增

sinx增区间是(2kπ-π/2,2kπ+π/2)
则2kπ-π/2<2x-π/4<2kπ+π/2
2kπ-π/4<2x<2kπ+3π/4
kπ-π/8所以减区间是(kπ-π/8,kπ+3π/8)

回答3:

π/2+2kπ≤2x-π/4≤3π/2+2kπ
3π/8+kπ≤x≤7π/8+kπ