Discrete time domain, Is cos(2n) periodic?
Consider a discrete-time sequence x[n] based on a sinusoid with angular frequency
\$\omega _0\$: x[n]=cos[\$\omega _0\$n]. 

If this sequence is periodic with a period of N samples, then the following must be true: cos[\$\omega _0\$ (n + N)] = cos [\$\omega _0\$n] (Eq. 1).                                         However, the left hand side can be expressed as cos [\$\omega _0\$(n + N)] = cos [\$\omega _0\$n + \$\omega _0\$N]. The cosine function is periodic with a period of 2\$\pi\$ and therefore cos [\$\omega _0\$n] = cos [\$\omega _0\$n + 2\$\pi\$r] (Eq.2), for integer values of r. Comparing Eq.1 and Eq.2 we get: 

$$\omega _0 N =2\pi ; \ \text{(for r=1)}$$ 
$$2\pi f _0 N =2\pi$$
$$f _0 = \frac{1}{N}$$

Since N is integer the signal will be periodic if \$f_0\$ is a rational number. Otherwise, it will be non-periodic. 

In your example, \$\omega _0\$ = 2 so \$f_0\$ = \$\frac{2}{2\pi}\$. But you cannot find an integer division (\$\frac{1}{N}\$) that results in an irrational number (\$f_0\$). So no N can be found to meet the criteria for this signal to be periodic.