已知集合A={α|2kπ≤α≤π+2kπ,k∈Z},B={α|-4≤α≤4},求A∩B.

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已知集合A={α|2kπ≤α≤π+2kπ,k∈Z},B={α|-4≤α≤4},求A∩B.
已知集合A={α|2kπ≤α≤π+2kπ,k∈Z},B={α|-4≤α≤4},求A∩B.

已知集合A={α|2kπ≤α≤π+2kπ,k∈Z},B={α|-4≤α≤4},求A∩B.
已知集合A={α|2kπ≤α≤π+2kπ,k∈Z},B={α|-4≤α≤4},求A∩B.
当k≥1时,2π>4,A∩B为空集
当k=0时,A={α|0≤α≤π}真包含于B={α|-4≤α≤4},
当k=-1时,A={α|-2π≤α≤-π},
A∩B={α|-4≤α≤-π},
当k≤-1时,-2π<-4,A∩B为空集
所以A∩B={α|-4≤α≤-π}U{α|0≤α≤π}={α|-4≤α≤-π或0≤α≤π}

A∩B.={α|-4≤α≤-π}∪{α|0≤α≤π} ( k取 -1或0)