Question

求助高一函数！！！：求函数y=3sin(2x+π/4),x∈〔0，π〕的单调递减区间。

Answer 1

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 2

0<=x<=π
0<=2x<=2π
π/4<=2x+π/4<=9π/4

sinx的减区间是(2kπ+π/2,2kπ+3π/2)
即第23象限
则次数是π/2<2x+π/4<3π/2
π/4<2x<5π/4
π/8<x<5π/8
所以减区间(π/8,5π/8)

Answer 3

求此函数的递减区间 就是求sin函数的递减区间
2x+π/4∈〔π/2+2kπ，3π/2+2kπ〕
求的x的范围 再与0，π〕求交集
所得的最终范围（π/8，5π/8〕

Answer 4

2x+π/4=π/2+2kπ x=π/8+kπ
2x+π/4=3π/2+2kπ x=5π/8+kπ