When do we divide by $2\pi$ while solving Nyquist problems?
Well!  all signals in this world are made up of sum of different rotations(sinusoidals) - different in three senses:
a. how big is the amplitude (A)
b. how fast is the rotation ($\omega$)
c. where is the starting point of the rotation (phase $\phi$)
Fourier made this very clear.
How do we measure rapidness of the rotating signals(sinusoidals) : by their angular velocity $\omega$ which is given in radians/seconds. This is the correct and most appropriate unit for measuring rotations - how much of angle is being covered per unit time! If a constituent signal is rotating by $200\pi$ radians in 1 second, we say $\omega = 200\pi$ rad/sec.
But we can also measure the rapidness of these rotating signals as number of rotations per unit time. That is $f$ expressed in $Hertz$ : number of rotations per second.
How are $f$ and $\omega$ related? One complete rotation covers a full circle meaning an angle of $360^o = 2\pi \ radians$. That means if the signal is making $200$ rotations per second then it is covering an angle of $200*2\pi = 400\pi$ per second. So, the relation between $f$ and $\omega$ is basically:
$$\omega = 2\pi f$$
That is why you divide $\omega$ by $2\pi$ when you want to express the frequency in $Hz$.
The example you have given is $\cos{\frac{\pi t}{3}} = \cos{\frac{2\pi t}{6}}$. The period of this sinusoidal is $T = 6$sec, therefore, the frequency in $Hz$ will be $f = \frac{1}{6}$ and in radians/sec will be $\omega = 2\pi f = \frac{2\pi}{6}$ rad/sec.
(Think about why the period of this sinusoidal is $6sec$. Figure out that the sinusoidal will repeat after $t = 6sec$. Figure out that the sinusoidal makes 1 full rotation of angle $2\pi$radians in $6sec$.)
And the sampling frequency is twice the frequency of the sinusoidal giving :
$$f_{sampling} = \frac{2}{6} = \frac{1}{3}sec$$
Meaning Sampling period is $\frac{1}{f_{sampling}} = 3sec$.