已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cosx+a,

已知函数f(x)=sin(2x+π/6)+sin(2x-π/6)+cosx+a,
,1.求最小正周期
2.求f(x)的单调增区间
3 求f(x)的最大值及此时x值
4.若当x∈【0,2/派】时,f(x)的最小值为2,求a
KERENYIN 1年前 已收到1个回答 举报

yueyue7577 幼苗

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f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a
= sin2xcosπ/6+cos2xsinπ/6 + sin2xcosπ/6-cos2xsinπ/6 + cos2x + a
= 2sin2xcosπ/6 + cos2x + a
= √3sin2x+cos2x+a
= 2(sin2xcosπ/6+cos2xsinπ/6) + a
= 2sin(2x+π/6) + a
最小正周期 = 2π/2 = π
2x+π/6∈(2kπ-π/2,2kπ+π/2)时单调增,∴单调递增区间(kπ-π/3,kπ+π/6),其中k∈Z
最大值2+a,此时2x+π/6=2kπ+π/2,即x=kπ+π/6,其中k∈Z
x∈[0,π/2]时,2x+π/6∈[-π/6,5π/6],2x+π/6=-π/6时最小值2sin(-π/6)+a=-1+a=2,a=3

1年前

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