Question

已知向量a=(根号3sin2x,cos2x),b=(cos2x,-cos2x).

1) x∈(7/24∏,5/12∏),a*b+1/2=-3/5,求cos4x
(2)cosx≥1/2,x∈(0,π),若关于x的方程a·b+1/2=m有且仅有一个实根，求实数m的值.

Answer 1

解：告唤(1)a*b+1/2=√3sin2xcos2x -(cos2x)^2+1/2
=√3/2sin4x - (1+cos4x )/2+1/2
=√3/2sin4x - cos4x /2
=-3/5
联立√3/2sin4x - cos4x /2 =-3/5，(sin4x)^2+(cos4x)^2=1
结合|cos4x|<=1得cos4x=(3-4√3 )/10
(2)a*b+1/2=√3sin2xcos2x -(cos2x)^2+1/2
=√3/2sin4x - (1+cos4x )/2+1/2
=√3/2sin4x - cos4x /2
=sin(4x-30°)
因为cosx≥1/2,x∈(0,π),
所以0<x<60°
所轿野以-30°<4x-30<210°
因为闭友喊a*b+1/2=m
sin(4x-30°)=m
所以-1/2<m<=0