Question

已知函数f(x)=2cos^2wx/2+cos(wx+π/3)的最小正周期为π

（一）求w的值
（二）在锐角三角形ABC中，abc分别为角ABC的对边，若f（a）=-1/2,c=3,三角形ABC的面积为3√3,求a的值
要过程啊~

Answer 1

f(x)=2cos^2wx/2+cos(wx+π/3)
=1+cos(wx)+cos(wx+π/3)
=1+2cos(wx+π/6)cos(-π/6)
=1+√3cos(wx+π/6)
最小正周期为T=2π/w=π => w=2

Answer 2

f(x)=2cos^2wx/2+cos(wx+π/3)
=1+cos(wx)+cos(wx+π/3)
=1+2cos(wx+π/6)cos(-π/6)
=1+√3cos(wx+π/6)
最小正周期为T=2π/w=π => w=2