Question

数学超难题

已知向量A=（2cosX，1），向量B=（cosX，√3sin2X）（X∈R），定义函数f（X）=向量A×向量B，若f（x）=1-√3，且x属于闭区间-π/3到π/3，则x=

Answer 1

f(x)=2cos²x+√3sin2X
=√3sin2X+cos2x+1
=2sin(2x+π/6)+1=1-√3
sin(2x+π/6)=-√3/2
x属于闭区间-π/3到π/3
2x+π/6属于闭区间-π/2到5π/6
所以 2x+π/6=-π/3
2x=-π/2
x=-π/4

Answer 2

向量A*向量B =2(cosx)^2-√3si2x
=1+cos2x-√3sin2x
=1+2cos(π/3+2x)=1-√3
cos(π/3+2x)=-√3/2
cos ( π-(π/3+2x))=√3/2
cos(2π/3-2x)=cosπ/3 或 cos(2π/3-2x)=cos(-π/3)
x=π/6 或 x=π/2(舍去）

Answer 3

x=-π/4

Answer 4

-π/6

Answer 5

-π/6