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已知函数f(x)=2根号3sinxcosx+2cos方x-1(x属于R) 求函数f(x)的单调递减区间
人气:385 ℃ 时间:2019-08-19 00:27:13
解答
f(x)=2√3sinxcosx+2cos²x-1=√3sin2x+cos2x=2sin(2x+π/6),则递减区间是:2kπ+π/2≤2x+π/6≤2kπ+3π/2,得:kπ+π/6≤x≤kπ+2π/3,则减区间是:[kπ+π/6,kπ+2π/3],其中k∈Z
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