求函数y=cos（-2x+π/3）的单调增区间
问题描述：
求函数y=cos（-2x+π/3）的单调增区间
答
2kπ-π≤-2x+π/3≤2kπ
2kπ-(4π/3)≤-2x≤2kπ-π/3
-kπ+(π/6)≤x≤-kπ+(2π/3)
∴函数单调增区间为
[-kπ+(π/6), -kπ+(2π/3)] k∈Z.
答
y=cos(2x-π/3)
∵y=cost的单调区间为[-π+2kπ,2kπ]为增，[2kπ,π+2kπ]为减
∴f(x)在[-π/3 +kπ,π/6+kπ]为增，[π/6+kπ，2π/3+kπ]为减。
答
2kπ-π≤-2x+π/3≤2kπ
2kπ-4π/3≤-2x≤2kπ-π/3
kπ+π/6≤x≤kπ+2π/3
答
cos(-2x+π/3)=cos(2x-π/3),
令2kπ+π≤2x-π/3≤2kπ+2π,得kπ+2π/3≤x≤kπ+7π/6
单调递增区间为[kπ+2π/3,kπ+7π/6](k∈Z)