Understanding Resonant Frequency in a Series RLC Circuit
resonance is when Xl = Xc


No, at series resonance X\$_L\$ = -X\$_C\$. This mathematically (and practically) means that those impedances cancel leaving just the series resistance, R at resonance.


  My reasoning behind this was that resonance is when Xl = Xc, meaning
  the circuit it purely resistive so the resistance would be at it's
  peak.


No, resistance remains exactly the same\$^1\$ for any frequency.


  I'm sure there is a simple explanation as to why, but I simply cannot
  get my head around it right now


Despite the two reactances cancelling out at resonance they are still their individually and you can still measure a voltage across each one but, as said before, the reactances are oppositely signed and, given that the current is common to all three components, the voltage across C and the voltage across L MUST be equal and opposite at series resonance.

Imagine a 1 ohm resistor in series with two 9V batteries wired in opposite directions. You put that in a black box and try and determine what's in the box: -







simulate this circuit &ndash; Schematic created using CircuitLab

All you will ever detect is a 1 ohm resistor.


  Which method is better? Are the reasonings for using either method
  correct?


The reasons are the same for both methods and both methods are equally good.



\$^1\$ Strictly speaking, as frequency rises so does the resistance of interconnecting leads (due to skin effect) but, for the sake of this answer, that is ignored.


  I used the equation 1/T, with T being the time period between the
  start of the wave and the peak of the wave. this gave me Wn (or Wo,
  whatever your preference).


You may be misled here - you should use the time to complete one full cycle, then take the reciprocal to find frequency (F) in hertz. If you desire \$\omega\$ then that is 2\$\pi F\$.