cutoff frequency of low pass filter with two resistors (differential signal)
Whenever you have a symmetrical circuit you can cut it in half to make the analysis easier. The equivalent half-circuit will show the same behavior. Please note that I am assuming that R2=R3.

In this case the symmetry might not be apparent, but by replacing the capacitor C1 by two series connected capacitors, each one having twice the value, the circuit becomes symmetrical. And the half circuit is a low-pass with R2 and 2*C that determine the corner frequency. Hence, $$f=\frac 1{2\pi\cdot R_2 \cdot  2C}$$
which is equivalent to your first equation, for R2=R3.