Don't power them from 3.3v because the forward voltage will vary with temperature of the leds.
The easiest and cheapest way would be to use some diodes to lower the 5v voltage closer to 3.6v and then use the resistor to limit the current.
For example, get some 2-3A diodes with forward voltage of around 1v ... ex
RL255 or
1n5408 and these will have around 0.8v..0.9v voltage drop at 1A.
If you don't have these, you could parallel two 1n400x diodes so that each diode will only transfer half an amp.
Anyway, with a diode (or two paralleled) in series with the 5v output, you'll have approx. 4.1..4.2v
Let's say you have a 3.6v 3W (3/3.6 = 0.83A) led ... now your formula to calculate resistor becomes a bit easier :
4.2v - 3.7v = 0.83A x R => R = (4.2-3.7) / 0.83 = 0.60 ohm ... so you could use 0.68 ohm, or two 1.2 ohm resistors in parallel.
The resistor will dissipate P = IxIxR = 0.83 x 0.83 x 0.6 = 0.41334 watts, so you'll have to use resistor (or resistors) rated for 1w
+5v ----> [ diode(s) >| ---- [ resistor(s) ]----- led +
-GND --------------------------------------------------led -
HOWEVER, what I would suggest is to get THREE of those leds and wire them in series, and you'll get 3 x 3.6v = ~ 10.8v and then you can power it from 12v
12v - 10.8v = 0.83A x R => R = 1.2v / 0.83A = 1.44 ohm ... so you can just use a 1.5 ohm resistor, or 2 x 3.3 ohm in parallel for 1.55 ohm (higher resistance means a bit less current)
The power dissipated will be 0.83 x 0.83 x 1.5 = 1.03335 watts, so you'd want to use a resistor rated for 3w or something like that.
ANother thing I'd suggest is to not expect those leds to be 3W... expect more like 1W and limit the current accordingly. Also... you MUST use a heatsink for the leds, or they'll die soon.