$ x(t)=e^{2jt} \to sys \to y(t)=te^{-2jt} $
$ x(t)=e^{-2jt} \to sys \to y(t)=te^{2jt} $
We want to know the output associated with the input $ x(t)=cos(2t) $. If you expand $ cos(2t) $ into two exponentials you will get $ \frac{e^{2jt} + e^{-2jt}}{2} $. Now you can use linearity to solve the problem. Linearity implies that if $ x(t)=cos(2t) $ then $ y(t)=tcos(-2t)=tcos(2t) $.
