f(x)=1/2*(1-cos2x)+1/2sqrt(3)sin2x-1/2 =1/2[sqrt(3)sin2x-cos2x] =sin(2x-pi/6) f(pi/12)=sin0=02x-pi/6=pi/2 x=pi/3x取pi/3时f(x)取最大值1sqrt（3）表示根号3
