设函数f(x)=向量a点乘向量b,向量a=(cosx,-√3),向量b=(2cosx,sin2x),若f(x)=1+√3,且x属于[-π/3,π/3],求x

题目

设函数f(x)=向量a点乘向量b,向量a=(cosx,-√3),向量b=(2cosx,sin2x),若f(x)=1+√3,且x属于[-π/3,π/3],求x

答案

∵a·b=2*(cosx)^2-√3*sin2x
=cos2x-sin2x+1
=2cos(2x+π/3)+1=1+√3.
∴cos(2x+π/3)=√3/2.
∵x∈[-π/3,π/3]
∴2x+π/3∈[-π/3,π]
∴2x+π/3=π/6或-π/6
则可得x=-π/12或-π/4.