②③
分析:①函数f(x)=cos
2x-cos
4x,满足f(x)=f(-x),f(x+

)=f(x);
②f(x)=sin
4x-cos
4x=sin
2x-cos
2x=-cos2x,满足f(x)=f(-x),f(x+

)=-f(x);
③

=

=-

cos2x,由②知符合条件;
④f(x)=|tanx|,满足f(x)=f(-x),f(x+

)=|cotx|≠-f(x),由此可得结论.
解答:①函数f(x)=cos
2x-cos
4x,满足f(x)=f(-x),f(x+

)=sin
2x-sin
4x=sin
2xcos
2x,f(x)=cos
2xsin
2x,∴f(x+

)=f(x),故①不符合;
②f(x)=sin
4x-cos
4x=sin
2x-cos
2x=-cos2x,满足f(x)=f(-x),f(x+

)=-cos(2x+π)=cos2x=-f(x),故②符合;
③

=

=-

cos2x,由②知符合条件;
④f(x)=|tanx|,满足f(x)=f(-x),f(x+

)=|cotx|≠-f(x),故④不符合
综上知,符合已知条件的函数序号为②③
故答案为:②③
点评:本题考查新定义,考查三角函数的化简,解题的关键是一一验证,属于中档题.