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已知函数f(x)=2根号3sinxcosx+2cos^2x-1,求f(x)的单调递减区间
人气:414 ℃ 时间:2019-08-20 21:48:18
解答
f(x)=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
令2x+π/6=z
sinz的单调递减区间为[2kπ+π/2,2kπ+3π/2](k∈Z)
∴2kπ+π/2≤2x+π/6≤2kπ+3π/2
∴kπ+π/6≤x≤kπ+2π/3 (k∈Z)
∴f(x)在区间[kπ+π/6,kπ+2π/3](k∈Z)是单调递减
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