函数y=cos(x/2-π/3)的单调递减区间是
问题描述：
函数y=cos(x/2-π/3)的单调递减区间是
答
方法一：
求导：y‘=-sin(x/2-π/3)*1/2=-1/2*sin(x/2-π/3)
∵单调递减 ∴y'=-1/2*sin(x/2-π/3)∴x/2-π/3∈(2kπ,π+2kπ)
得出：2π/3+4kπ
∵y=cos(x/2-π/3)是周期函数且T=2π
∴2kπ得出2π/3+4kπ
答
求导：y‘=-sin(x/2-π/3)*1/2=-1/2*sin(x/2-π/3)
∵单调递减 ∴y'=-1/2*sin(x/2-π/3)∴x/2-π/3∈(2kπ,π+2kπ)
得出：2π/3+4kπ
答
y'=-2sin(2x-π/3),若函数单调递减,应有
y'=-2sin(2x-π/3)0,0
答
余弦函数的单调减区间是[2kπ,2kπ+π]
所以2kπ4kπ+2π/3即原函数单调减区间为[4kπ+2π/3, 4kπ+8π/3]