已知函数f（x）=sin（x+π/6)+cos(x+π/3)+sinx+a(a属于R,且a为常数),若函数f(x)在[-π/2,π/2]上的最大值

f（x）=sin（x+π/6)+cos(x+π/3)+sinx+a=sinx*cosπ/6+cosx*sinπ/6+cosx*cosπ/3-sinx*sinπ/3+sinx+a=√3/2sinx+1/2cosx+1/2cosx-√3/2*sinx+sinx+a=cosx+sinx+a=√2sin(x+π/4)+af(x)在[-π/2,π/2]上的最大值1+...