Question

求函数y=3sin（2x+π/4），x∈[0，π]的单调递减区间

Answer 1

y=3sin（2x+π/4）
减区间

令π/2+2kπ<=2x+π/4<=3π/2+2kπ,k∈Z
π/8+kπ<=x<=5π/8+kπ,k∈Z
∵x∈[0，π]
∴令k=0
x∈[π/8,5π/8]
单调递减区间[π/8,5π/8]
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Answer 2

2X+π/4∈[0，5/4π]，因为正弦函数的单调递减区间为x∈[π/2+2kπ，3π/2+2kπ]，所以π/2+2kπ<2X+π/4<3π/2+2kπ,所以X∈[π/8+kπ，5π/8+kπ].

Answer 3

解：sinx在(π/2+2kπ，3π/2+2kπ)之间是单调递减的！
所以
把2x+π/4看做一个整体可以知道
π/2+2kπ
<
(2x+π/4)<3π/2+2kπ
解出
单调递减区间为
π/8+kπ<x<5π/8+kπ

Answer 4

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 5

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