已知函数f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2
⑴求f(x)的最小正周期⑵求f(x)的最小值及此时x的值
人气:344 ℃ 时间:2020-06-02 22:50:40
解答
sin(2x+π/3)-√3sin²x+sinxcosx+√3/2
=sin(2x+π/3) - 2sinx【(√3/2)sinx -(1/2)cosx】+(√3/2)
=sin(2x+π/3) - 2sinx【sin(π/3)sinx - cos(π/3)cosx】+(√3/2)
=sin(2x+π/3) + 2 sinx cos(x+π/3)+(√3/2)
=sin(2x+π/3) + 【sin(2x+π/3)+sin(π/3) 】+(√3/2)
=2sin(2x+π/3) +(√3/2)+(√3/2)
=2sin(2x+π/3) + √3
最小正周期为π
最小值为-2+√3,此时的 x =(k+7/12)π
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