Question

求下列函数的单调区间y=sin(x+π/3) y=cos2x

。。。。。

Answer 1

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+π/2, => 2kπ-5π/6<x<2kπ+π/6,
则y=sin(x+π/3) 在[2kπ-5π/6 , 2kπ+π/6]单调增加，在( 2kπ+π/6, 2kπ+π7/6)单调减小，k∈Z
2)y=cos2x
=> y'=-2sin2x,
=> y'>0时，sin2x<0,=> 2kπ-π<2x<2kπ, => kπ-π/2<x<kπ,
则y=cos2x在[kπ-π/2,kπ]单调增加，在(kπ,kπ+π)单调减小，k∈Z

Answer 2

(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

Answer 3

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]