Question

求函数Y=3sin（2x+π/4），x属于[0，π]的单调递减区间。

Answer 1

因为y=sinx的单调递减区间是[π/2+2kπ,3π/2+2kπ] ，(k∈Z+)
π/2+2kπ≤2x+π/4≤3π/2+2kπ
2kπ+π/4≤2x≤2kπ+5π/4
kπ+π/8≤x≤kπ+5π/8
又因为x∈[0，π]
所以 y=3sin(2x+π/4)，x属于[0，π]的单调递减区间是 [π/8,5π/8]

Answer 2

2x+π/4=π/2+2kπ x=π/8+kπ
2x+π/4=3π/2+2kπ x=7π/8+kπ

函数y=3sin(2x+π/4),x∈〔0，π〕的单调递减区间
[π/8+kπ,7π/8+kπ],k为整数

Answer 3

y=sinx的单调递减区间是[π/2+2kπ,3π/2+2kπ] ，(k∈Z+)
π/2+2kπ≤2x+π/4≤3π/2+2kπ ，(k∈Z+)
kπ+π/8≤x≤kπ+5π/8，(k∈Z+)
所以 y=3sin(2x+π/4)在x属于[0，π]的单调递减区间是 [π/8,5π/8]