已知向量m=(sinx,-1),向量n=(根号3cosx,-1/2),函数f(X )=m^2+m*n-2.求f（x）的最大值

f(X )=m^2+m*n-2=sin^2x+1+ 根号3sinxcosx+1/2=sin^2x+ 根号3sinxcosx+3/2 =1/2(2sin^2x-1)+ 根号3sinxcosx+2=根号3/2sin2x-1/2cos2x+2=sin(2x-60)+2 1<=sin(2x-60)+2 <=3 最大值3