This problem uses the linearity. If we know that:
$ e^{2jt} \to te^{-2jt} $ and $ e^{-2jt} \to te^{2jt} $Then by rewriting cos(2t) as $ \frac{e^{2jt} + e^{-2jt}}{2} $ then since the system in linear, take $ \frac{1}{2}e^{2jt} + \frac{1}{2}e^{-2jt} $ through the system to get $ \frac{1}{2}te^{-2jt} + \frac{1}{2}te^{2jt} $ which is the same as $ t\cos(2t) $
