求函数y=3sin(派/6-x/2)-1的单调区间

Question

如题所述

Answer 1

y = 3sin(π/6 – x/2) – 1,令t = -x/2,t在x∈R上关于x单调递减.而y = 3sin(π/6 + t) – 1,在(π/6 + t)∈[2kπ – π/2,2kπ + π/2],即t∈[2kπ – 2π/3,2kπ + π/3],k∈Z上关于t单调递增；y在(π/6 + t)∈[2kπ + π/2,2kπ + 3π/2],即t∈[2kπ + π/3,2kπ + 4π/3],k∈Z上关于t单调递减；把t = -x/2代入上述关系,联系t关于x在R上递减可得原函数的单调递增区间为[-8π/3 – 4kπ,-2π/3 – 4kπ],k∈Z；单调递减区间为[-2π/3 – 4kπ,4π/3 – 4kπ],k∈Z .