1、
由x范围则cosx>0
sin²x+cos²x=1
所以cosx=3/5
所以f(x)=(4√3-3)/5
2、
f(x)=2(sinx*√3/2-cosx*1/2)
=2(sinxcosπ/6-cosxsinπ/6)
=2sin(x-π/6)
-π/2<x<π/2
-2π/3<x-π/6<π/3
所以x-π/6=π/3
即x=π/2,最大值=sinπ/3=√3/2