Question

求出数 y=sin(π/3-1/2x) x∈[-2π，2π]的单调递增区间

求出数 y=sin(π/3-1/2x) x∈[-2π，2π]的单调递增区间
很急 正确点的过程
谢谢

Answer 1

y=-sin(x/2-π/3)
所以y增区间即sin(x/2-π/3)的减区间
sinx的减区间是(2kπ+π/2,2kπ+3π/2)

2kπ+π/2<x/2-π/3<2kπ+3π/2
2kπ+5π/6<x/2<2kπ+11π/6
4kπ+5π/3<x<4kπ+11π/3

k=-1,则-7π/3<x<-π/3
和[-2π,2π]交集是[-2π,-π/3)

k=0,则5π/3<x<11π/3
和[-2π,2π]交集是(5π/3,2π]

所以y的增区间是(-2π,-π/3)∪(5π/3,2π)

Answer 2

sinx的递增区间是 X∈[-0.5π+2Kπ，0.5π+2Kπ]
然后解出X