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

Question

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.

Answer 1

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