已知向量a=（2sinx,cosx),向量b=（√3cosx,2cosx),函数f(x)=2sin(2x+π/6)
问题描述：
已知向量a=（2sinx,cosx),向量b=（√3cosx,2cosx),函数f(x)=2sin(2x+π/6)
若f(x1）=6/5,x1∈[π/4,π/2],求cos2x1
答
f(x)=2sin(2x+π/6)
f(x1）=6/5
2sin(2x1+π/6)=6/5
sin(2x1+π/6)=3/5
∵x1∈[π/4,π/2]
∴2x1∈[π/2,π]
∴2x1+π/6∈[2π/3,7π/6]
∴cos(2x1+π/6)=-4/5
∴cos2x1=cos[(2x1+π/6)-π/6]
=cos(2x1+π/6)cosπ/6+sin(2x1+π/6)sinπ/6
=-4/5*√3/2+3/5*1/2
=(3-4√3)/8