Comparing 2 sinusoids: Discrete case vs Continuous case I'm doing MIT's OCW course with Allan Oppenheim - "Signals and Systems".

I'm given 2 continuous signals$$x(t)=\cos[\omega_x(t+r_x)+\theta_x]\tag1$$ and $$y(t)=\cos[\omega_y(t+r_y)+\theta_y]\tag2$$

Assuming their phases are the same, the notes tell me that the signals are identical if $$\omega_x=\omega_y$$which is quite obvious.

Now, the discrete case:

Again, I'm given 2 signals $$x[n]=\cos[\Omega_x(n+P_x)+\theta_x]\tag3$$ and $$y[n]=\cos[\Omega_y(n+P_y)+\theta_y]\tag4$$

This time, assuming that the phases are the same, the notes say that the signals are identical if $$\Omega_x=\Omega_y+2\pi k \tag5$$

Why is the $2\pi k$ present in equation 5?
Because of the $2\pi$-periodicity of the cosine and the discrete nature of $n$ ($n,k\in \mathbb{N}$):
$$
x[n] = \cos[\Omega_\mathrm{x}] = \cos[(\Omega_\mathrm{y}+2\pi k)n] = \cos[\Omega_\mathrm{y}n+2\pi kn] = \cos[\Omega_\mathrm{y}n] = y[n]
$$
This important "phenomenon" is called aliasing and only occurs for discrete-time signals. The normalized frequency $\Omega$ is therefore usually restricted to the interval $[-\pi, \pi]$.