Relation between CTFT and DTFT I need to transform this function:

$$
x(t) = 4\sin(20\pi t ) - 5\cos(24\pi t ) + 3\sin(120\pi t )
$$ 
into a sequence $x(n)$ given that the sampling frequency should be 50 Hz. So that means the sampling interval $T = \frac{1}{50} = 0.02\ s\ $. So would the sequence be:

$$
x[n] = 4\sin(20\pi0.02n ) - 5\cos(24\pi 0.02n ) + 3\sin(120\pi 0.02n)
$$

I dont know how to compare the CTFT of $x(t)$ and the DTFT of $x(n)$. Any idea ?
Your sequence simply is a sampled version of the continuous signal $x(t)$. So

$$x_d(n)=x(nT),\quad T=\frac{1}{50}$$

The relation between the CTFT of $x(t)$ and the DTFT of $x_d(n)$ is that the DTFT is the sum of shifted versions of the CTFT. The spectra are shifted by the sampling frequency. That's the reason for aliasing.