函数y=cos(x/2-π/3)的单调递减区间是

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函数y=cos(x/2-π/3)的单调递减区间是

函数y=cos(x/2-π/3)的单调递减区间是
函数y=cos(x/2-π/3)的单调递减区间是

函数y=cos(x/2-π/3)的单调递减区间是
y'=-2sin(2x-π/3),若函数单调递减,应有
y'=-2sin(2x-π/3)0,0

余弦函数的单调减区间是[2kπ,2kπ+π]
所以2kπ<= x/2-π/3<=2kπ+π
4kπ+2π/3<=x<=4kπ+8π/3
即原函数单调减区间为[4kπ+2π/3, 4kπ+8π/3]

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)<0
∴x/2-π/3∈(2kπ,π+2kπ)
得出:2π/3+4kπ

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)<0
∴x/2-π/3∈(2kπ,π+2kπ)
得出:2π/3+4kπ方法二:
∵y=cos(x/2-π/3)是周期函数且T=2π
∴2kπ< x/2-π/3<2kπ+...

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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)<0
∴x/2-π/3∈(2kπ,π+2kπ)
得出:2π/3+4kπ方法二:
∵y=cos(x/2-π/3)是周期函数且T=2π
∴2kπ< x/2-π/3<2kπ+π
得出2π/3+4kπ一般情况下,单调区间取开区间即可

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