由題意知f(x)=f(-x)
所以sin(x+a)+√3·cos(x-a)=sin(-x+a)+√3·cos(-x-a)
sin(x+a)+√3·cos(x-a)=-sin(x-a)+√3·cos(x+a)
√3·cos(x-a)+sin(x-a)=√3·cos(x+a)-sin(x+a)
√3/2·cos(x-a)+1/2·sin(x-a)=√3/2·cos(x+a)-1/2·sin(x+a)
sinπ/3 ·cos(x-a)+cosπ/3·sin(x-a)=sinπ/3·cos(x+a)-cosπ/3·sin(x+a)
sin(π/3+x-a)=sin(π/3-x-a)
sin(π/3+x-a)-sin(π/3-x-a)=0
2cos[(π/3+x-a+π/3-x-a)/2]·sin[(π/3+x-a-π/3+x+a)/2] =0
cos(π/3-a)·sinx=0
上式無(wú)論x取任何數(shù)都成立,所以cos(π/3-a)=0
π/3-a=-kπ-π/2,a=kπ+5π/6