What we know
The two things that are important when solving this problem is that we know what the output is for $ e^{j2t} $ and $ e^{-j2t} $ and that the system is linear.
$ e^{j2t}\! $ ---> system ---> $ te^{-j2t}\! $
$ e^{-j2t}\! $ ---> system ---> $ te^{j2t}\! $
What we can do with this
First step is to take the input $ cos(2t) $ and relate it to the two known inputs. It is easy to do this using Euler's Identity $ e^{jat} = \frac{cos(t) + jsin(t)}{a} $
$ cos(2t) = \frac{cos(2t) + jsin(2t)}{2} + \frac{cos(2t) - jsin(2t)}{2} = \frac{e^{j2t} + e^{-j2t}}{2} $
So now that we have $ cos(t) $ represented as a sum of our two outputs we can run those through the system.
$ .5e^{j2t}+.5e^{-j2t}\! $ ---> system ---> $ .5te^{j2t}+.5te^{-j2t}\! $
Now, we can use Euler's Identity again to get our answer in terms of cosine.
$ t\frac{e^{j2t}+e^{-j2t}}{2} = tcos(2t) $
