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求下列函数的单调区间y=sin(x+π/3) y=cos2x
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人气:279 ℃ 时间:2020-04-12 16:04:38
解答
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-5π/60时,sin2x 2kπ-π
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