If:
$\displaystyle H(\omega)=\frac{1}{1+j\omega}$
then:
$\displaystyle H(s)=\frac{1}{1+s}$
and:
$\displaystyle Y(s)=\frac{1}{1+s}X(s)$
or:
$\displaystyle (1+s)Y(s)=X(s)$
which is equivalent to the differential equation in the time domain:
$\displaystyle \left[ \frac{d}{dt}+1 \right]y(t)=x(t)$
or:
$\displaystyle y'+y=x$
So putting $\displaystyle x(t)=\cos(t)$ we have:
$\displaystyle y'(t)+y(t)=\cos(t)\ \ \ \dots[1]$
Now how you solve this depends on what you know.
One approach is, as a LTI system driven by a sinusoid has an sinusoidal output at the same frequency as the input, use a trial solution of the form $\displaystyle y(t)=A\sin(t)+B\cos(t)$.
Substitute this into $\displaystyle [1] $above and solve for $\displaystyle A$ and $\displaystyle B$.
This isn't quite correct. Fourier Transforms imply steady state solutions whereas Laplace Transforms reveal transient behavior as well.
Simply what you want to do is set
$X(\omega) = \mathscr{F}\{\cos(t)\}$
$H(\omega) = \dfrac{1}{1+j \omega}$
$Y(\omega) = H(\omega)X(\omega)$
$y(t) = \mathscr{F}^{-1}\{Y(\omega)\}$
The steady state answer is that $y(t) = \dfrac{1}{2} (\sin (t)+\cos (t))$
Including transients, i.e. repeating above with Laplace transforms we get $y(t) = \dfrac{1}{2} (\sin (t)+\cos (t))-\frac{e^{-t}}{2}$
This second answer, including transients, is what is found by using the differential equation approach done by zzephod
The first solution is what you get from my method, as it assumes the output is sinusoidal which forces out the transient response.
We can do this because the input is $\displaystyle x(t)=\cos(t)$ not $\displaystyle x(t)=u(t)\cos(t)$. That is we have an infinite expanse of time before $\displaystyle t=0$ with a sinusoidal input, which is why the output is going to be sinusoidal. As I said it all depends on how much and what you know.