Question

求函数y=2sin(π/3-2x)的单调减区间是,x∈（0,2π）的单调区间.

Answer 1

y=2sin(π/3-2x)=- 2sin(2x-π/3),
求函数y=2sin(π/3-2x)的单调减区间即求函数2sin(2x-π/3)的递增区间.
2kπ-π/2≤2x-π/3≤2kπ+π/2,k∈Z.
kπ-π/12≤x≤kπ+5π/12,k∈Z.
x∈（0,2π）,k=0时,(0,5π/12]递减；
k=1时,[11π/12,17π/12] 递减.
k=2时,[23π/12,2π）递减.
求函数y=2sin(π/3-2x)的单调增区间即求函数2sin(2x-π/3)的递减区间.
2kπ+π/2≤2x-π/3≤2kπ+3π/2,k∈Z.
kπ+5π/12≤x≤kπ+11π/12,k∈Z.
x∈（0,2π）,k=0时,[5π/12,11π/12] 递增；
k=1时,[17π/12,23π/12] 递减.