求下列函数的单调区间y=sin(x+π/3) y=cos2x.
问题描述：
求下列函数的单调区间y=sin(x+π/3) y=cos2x
.
答
(1) y=sin(x+π/3)
单增区间为x+π/3∈[2kπ-π/2, 2kπ+π/2] x∈[2kπ-5π/6, 2kπ+π/6], k∈Z
单减区间为x+π/3∈[2kπ+π/2, 2kπ+3π/2] x∈[2kπ+π/6, 2kπ+7π/6], k∈Z
(2)y=cos2x
单增区间为2x∈[2kπ-π, 2kπ] x∈[kπ-π/2, kπ], k∈Z
单减区间为2x∈[2kπ, 2kπ+π] x∈[kπ, kπ+π/2], k∈Z
希望能帮到你，祝学习进步O(∩_∩)O
答
单增：[-5π/6+2kπ,π/6+2kπ] 单减：[π/6+2kπ,7π/6+2kπ]
单增：[π/2+kπ，π+kπ] 单减：[kπ,π/2+kπ]
答
1) y=sin(x+π/3)
=> y'=cos(x+π/3)
y'>0时,cos(x+π/3)>0,=> 2kπ-π/2＜x+π/3 2kπ-5π/60时,sin2x 2kπ-π