${\int}_{a}^{b}\mathrm{sin}(2\pi t)\cdot \mathrm{cos}(2\pi t)dt=\frac{\mathrm{cos}(4\pi b)-cos(4\pi a)}{8\pi}=0$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}(4\pi b+2\pi k)=cos(4\pi a+2\pi l),k,l\in \mathbb{Z}$

I don't know how to solve this for a and b, can anybody help me with that please?