How to get output from High Pass filter on this signal?
What you get as $x(\omega)$ is two $\mathrm{sinc}(\pi t) = \displaystyle\frac{\sin(\pi t)}{\pi t}$ shifted by $+\omega_0$ and $-\omega_0$ in frequency ($\omega_0 = 3 \pi$). By the frequency shift property of the FT:

$$ 
\mathcal F\left\{x(t) \cdot e^{i\omega_0 t}\right\} = X\left(e^{j(\omega - \omega_0)}\right)
$$

Then, remember $e^{i\omega t} = \cos(\omega t) + i \cdot \sin(\omega t)$. Therefore:

$$
x(\omega) = \sqcap_{2\pi} (\omega - \omega_0) + \sqcap_{2\pi} (\omega + \omega_0)
$$

Doing inverse:
$$
\begin{align}
x(t) &amp;= \mathcal F^{-1}\left\{x(\omega)\right\} \\
 &amp;=\mathrm{sinc} (\pi t) \cdot \big(\cos(\omega_0 t) + i \cdot \sin(\omega_0 t)\big) +  \mathrm{sinc} (\pi t) \cdot \big(\cos(-\omega_0 t) + i \cdot \sin(-\omega_0 t)\big) \\
&amp;= 2 \cdot \mathrm{sinc} (\pi t)\cdot \cos(\omega_0 t) \\
&amp;= 2 \cdot \frac{\sin(\pi t)}{\pi t}\cdot \cos(3 \pi t)
\end{align}
$$

since $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$