已知函数fx=根号3sin^2x+sinxcosx - 知识问答

已知函数fx=根号3sin^2x+sinxcosx,x属于[派/2,派],求fx零点

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f(x)=√3sin²x+sinxcosx
=√3[(1-cos2x)/2]+1/2sin2x
=1/2sin2x-√3/2cos2x+√3/2
=sin(2x-π/3)+√3/2
∵x∈[π/2,π],∴2x-π/3∈[2π/3,5π/3]
∴零点是使2x-π/3=π的点,解得x=2π/3
∴零点是2π/3

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