Question

y=sin(π/6-x)的单调递减区间为?

要过程

Answer 1

)∵Y=sin(π／6-X)=-sin(X-π／6)，
∴Y=sin(π／6-X)的单调递减区间
就是Y=sin(X-π／6)的递增区间，
由2kπ-π/2≤X-π／6≤2kπ+π/2（k∈Z）得
2kπ-π/3≤X≤2kπ+2π/3（k∈Z），
∴Y=sin(π／3-2X)的单调递减区间为
[2kπ-π/3，2kπ+2π/3]（k∈Z）.

Answer 2

因为y=sinx的单调递减区间为[π/2+2kπ,3π/2+2kπ](k为整数)；所以π/2+2kπ<=π/6-x<=3π/2+2kπ,解得 -4π/3+2kπ<=x<=-π/3+2kπ

Answer 3

因为y=sin(x)单调递减区间为(π/2+2kπ,3π/2+2kπ)所以π/2+2kπ<π/6-x<3π/2+2kπ即-4π/3-2kπ<x<-π/3-2kπ