Being irreducible is a much stronger property than having no roots (in the given field). Having no roots means that there are no linear factors. But for a polynomial to be irreducible it must have no nontrivial factors at all.
As Bruno J. points out, the roots of the given polynomial in the complex field are the complex 11th roots of unity. These can be grouped in complex conjugate pairs to form quadratic factors (for k = 1,2,3,4,5) over the real field. To show that the polynomial is irreducible over the rationals, you would have to show that none of these can be grouped together to form a rational polynomial. I don't know how to do that, but presumably it must involve showing that those cosines are seriously irrational.