已知向量m=(2cosx,根号3cosx-sinx),n=(sin(x+派/6),sinx),且满足f(x)=m·n.(1)求函数y=f(x)的单调递增区间;(2)设三角形ABC的内角A满足f(A)=2,a、b、c分别为角A、B、C所对的边,且向量AB·向量AC=根号3,求边BC的最
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![已知向量m=(2cosx,根号3cosx-sinx),n=(sin(x+派/6),sinx),且满足f(x)=m·n.(1)求函数y=f(x)的单调递增区间;(2)设三角形ABC的内角A满足f(A)=2,a、b、c分别为角A、B、C所对的边,且向量AB·向量AC=根号3,求边BC的最](/uploads/image/z/6969318-6-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fm%3D%282cosx%2C%E6%A0%B9%E5%8F%B73cosx-sinx%29%2Cn%3D%28sin%28x%2B%E6%B4%BE%2F6%29%2Csinx%29%2C%E4%B8%94%E6%BB%A1%E8%B6%B3f%28x%29%3Dm%C2%B7n.%281%29%E6%B1%82%E5%87%BD%E6%95%B0y%3Df%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%3B%282%29%E8%AE%BE%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E5%86%85%E8%A7%92A%E6%BB%A1%E8%B6%B3f%28A%29%3D2%2Ca%E3%80%81b%E3%80%81c%E5%88%86%E5%88%AB%E4%B8%BA%E8%A7%92A%E3%80%81B%E3%80%81C%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%2C%E4%B8%94%E5%90%91%E9%87%8FAB%C2%B7%E5%90%91%E9%87%8FAC%3D%E6%A0%B9%E5%8F%B73%2C%E6%B1%82%E8%BE%B9BC%E7%9A%84%E6%9C%80)
已知向量m=(2cosx,根号3cosx-sinx),n=(sin(x+派/6),sinx),且满足f(x)=m·n.(1)求函数y=f(x)的单调递增区间;(2)设三角形ABC的内角A满足f(A)=2,a、b、c分别为角A、B、C所对的边,且向量AB·向量AC=根号3,求边BC的最
已知向量m=(2cosx,根号3cosx-sinx),n=(sin(x+派/6),sinx),且满足f(x)=m·n.(1)求函数y=f(x)的单调递增区间;(2)设三角形ABC的内角A满足f(A)=2,a、b、c分别为角A、B、C所对的边,且向量AB·向量AC=根号3,求边BC的最小值.
已知向量m=(2cosx,根号3cosx-sinx),n=(sin(x+派/6),sinx),且满足f(x)=m·n.(1)求函数y=f(x)的单调递增区间;(2)设三角形ABC的内角A满足f(A)=2,a、b、c分别为角A、B、C所对的边,且向量AB·向量AC=根号3,求边BC的最
向量m=(2cosx,√3cosx-sinx),n=(sin(x+π/6),sinx),且满足f(x)=m·n
f(x)=m·n
=2√3sinxcosx+cos²x-sin²x
=√3sin2x+cos2x
=2sin(2x+π/6)
(1).由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z,得:
kπ-π/3≤x≤kπ+π/6,k∈Z,
∴(x)的单调递增区间为:[kπ-π/3,kπ+π/6],(k∈Z)
(2).∵f(A)=2sin(2A+π/6)=2
∴sin(2A+π/6)=1
又∵0
1、f(x)=m*n=(cosx+sinx)(cosx-sinx)+2√3sinxcosx=cos²x-sin²x+√3sin2x=cos2x+√3sin2x=2sin(2x+π/6),单调增区间:2kπ-π/2≤2x+π/6≤2kπ+π/2,得:kπ-π/3≤x≤kπ+π/6,增区间是[kπ-π/3,kπ+π/6],其中k是整数。
2、f(A)=1,得:sin(2A+...
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1、f(x)=m*n=(cosx+sinx)(cosx-sinx)+2√3sinxcosx=cos²x-sin²x+√3sin2x=cos2x+√3sin2x=2sin(2x+π/6),单调增区间:2kπ-π/2≤2x+π/6≤2kπ+π/2,得:kπ-π/3≤x≤kπ+π/6,增区间是[kπ-π/3,kπ+π/6],其中k是整数。
2、f(A)=1,得:sin(2A+π/6)=1,则A=π/6。因a/sinA=b/sinB=c/sinC,则:a/sinA=(b+c)/(sinB+sinC),代入,得:sinB+sinC=1,sinB+sin(120°-B)=1,sinB+(√3/2)cosB-(1/2)sinB=1,(1/2)sinB+(√3/2)cosB=1,sin(B+π/3)=1,得:B=30°,所以C=120°
3、根号3-1
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