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函数y=sin2x+根号3cos2x的单调区间
人气:256 ℃ 时间:2020-04-06 23:21:46
解答
y=sin2x+√3cos2x
=2sin(2x+π/3)
单调增区间:
令2kπ-π/2<2x+π/3<2kπ+π/2,k∈Z
得kπ-5π/12<x<kπ+π/12,k∈Z
所以函数的单调增区间是(kπ-5π/12,kπ+π/12),k∈Z
单调减区间:
令2kπ+π/2<2x+π/3<2kπ+3π/2,k∈Z
得kπ+π/12<x<kπ+7π/12,k∈Z所以函数的单调减区间是(kπ+π/12,kπ+7π/12),k∈Z
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