1)
f(x)=6cos^2(wx/2)+√3sinwx-3=3(1+coswx)+√3sinwx-3=√3sinwx+3coswx
f(x)=√3sinwx+3coswx=2√3[sinwxcos(π/3)+coswxsin(π/3)]=
f(x)=2√3sin(wx+π/3)
过A点作AA ' ⊥OX于X ' AA '=2√3 因为ΔABC是正三角形,所以AA '=√3BA '=2√3所以BA '=2
BA ' =(1/4)*T=2 ==>T=8=2π/w ==>w=π/4
f(x)=2√3sin[(π/4)x+π/3]
2)
2√3sin[(π/4)x0+π/3]=8√3/5 ==>sin[(π/4)x0+π/3]=4/5 ①
-10/3