Question

求助一道简单数学题

函数y=1/2sin(π/4-2/3x)的单调递增区间为？？？
要步骤！！谢谢

Answer 1

解
∵y=sinx的单调增区间是[-π/2+2kπ,π/2+2kπ](k∈Z)，
∴用上诉公式代入得-π/2+2kπ≤2/3x-π/4≤π/2+2kπ
解得：2kπ-π/4≤-2/3x≤2kπ+3π/4
9π/8+3kπ≥x≥3kπ+21π/8
所以
增区间(3kπ+9π/8,3kπ+21π/8)

Answer 2

y=-1/2sin(2/3x-π/4)
所以y和sin(2/3x-π/4)单调性相反

sinx的增区间是(2kπ-π/2,2kπ+π/2)
减区间是(2kπ+π/2,2kπ+3π/2)
所以sin(2/3x-π/4)单调增时
2kπ-π/2<2/3x-π/4<2kπ+π/2
2kπ-π/4<2/3x<2kπ+3π/4
3kπ-3π/8<x<3kπ+9π/8
同理，单调减时是3kπ+9π/8<x<3kπ+21π/8

y和sin(2/3x-π/4)单调性相反
所以
增区间(3kπ+9π/8,3kπ+21π/8)
减区间(3kπ-3π/8,3kπ+9π/8)