已知向量a=(sin2x,2sinx),b=(√3,-sinx),函数f(x)=a*b①求函数f(x)的最大值②求函数f
1个回答

(1) f(x)=ab=√3sin2x-2sin²x

=√3sin2x-(1-cos2x)

=2[(√3/2)sin2x+(1/2)cos2x]-1

=2sin(2x+π/6) -1

当sin(2x+π/6)=1时,f(x)有最大值为1

(2) 令f(x)=2sin(2x+π/6) -1=0,得

sin(2x+π/6) =1/2,

2x+π/6=π/6+2kπ或 2x+π/6=5π/6+2kπ,

解得 零点的集合为

{x|x=kπ或 x=π/3+kπ,k∈Z｝

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