fx=2根号3sinxcosx+2cos平方x-1+a

已知函数f(x)=2根号3sinxcosx+2cos^2x-1(x属于R),(1)求函数f(x)的单调递减区间,(2)若A是

第1个回答  2019-08-08
f(x)=2√3sinxcosx+2(cosx)^2-1
=√3sin2x+cos2x
=2(sin2xcosπ/6+cos2xsinπ/6)
=2sin(2x+π/6),
2kπ+π/2≤2x+π/6≤2kπ+3π/2,单调递减,k∈Z,
函数f(x)的单调递减区间:kπ+π/6≤x≤kπ+2π/3,k∈Z
∴x∈[kπ+π/6,kπ+2π/3],k∈Z.

fx=2根号3sinxcosx+2cos平方x-1+a
f(x)=2√3sinxcosx+2(cosx)^2-1 =√3sin2x+cos2x =2(sin2xcosπ\/6+cos2xsinπ\/6)=2sin(2x+π\/6),2kπ+π\/2≤2x+π\/6≤2kπ+3π\/2,单调递减,k∈Z,函数f(x)的单调递减区间:kπ+π\/6≤x≤kπ+2π\/3,k∈Z ∴x∈[kπ+π\/6,kπ+2π\/3],k∈Z.

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