I know you've selected an answer already. But I thought I'd add a more detailed (and exaggerated) picture of the behavior that captures all of what's happening. It may supplement this answer with added information that someone else may appreciate.
There are three time periods indicated on the diagram and two curves.
The orange curve is what's left of the original sine wave after the bridge rectifier diodes have had their voltage-dropping impact. If you look closely at the orange curve, you will see a short period of time where the value is zero and the sine/cosine shape isn't present, anymore. This is during a short time when the diodes are effectively off (or in the process of turning off.)
So, the orange curve is what you'd observe across your load resistor if there was no capacitor present.
The blue curve shows you what the RC decay curve might look like (the actual curve, with respect to the orange curve, depends on the values of R and C and also the AC frequency.)
A load resistor will draw a gradually declining current from the capacitor, when the blue curve is above the orange curve. The capacitor supplies the load current when the sinusoidal transformer voltage, less the diode drops, is below the capacitor's voltage because the diodes are reverse-biased then and no longer support any of the load current.
I've super-imposed these two curves so that you can see how they interact a little better. During period \$T_1\$, the transformer and bridge rectifier are supplying all of the load current as well as any current required by the capacitor (as the capacitor voltage also follows the orange curve.) During period \$T_2\$ the transformer and bridge rectifier are supplying some of the load current, less some current supplied by the capacitor as it discharges somewhat. But during \$T_3\$, the sinusoidal shaped result from the transformer and bridge rectifier have fallen below the capacitor voltage at a faster rate than the load resistor can drain the capacitor and so during \$T_3\$ the capacitor is supplying all of the load current by itself and the shape of the voltage across the load is then a simple RC decay shape.
If the resistor weren't present, this is the condition of "no load." What happens then is that the blue line stays flat and sits on the horizontal line labeled \$V_p\$ on the chart. In short, it rises to the peak voltage and just stays there (ignoring capacitor leakage current.)
So the voltage across the load resistor will follow the orange curve for the two periods, \$T_1\$ and \$T_2\$, and will follow the blue curve during \$T_3\$.
Keep in mind that I've exaggerated the decay curve (it usually isn't allowed to dive down as deeply as I've allowed in the graph) because I want to also highlight what's happening in period \$T_2\$, which is a period that is all too rarely shown in graphs you will more easily find on the web. The period \$T_2\$ is usually less than 1% of the total period and most people reasonably choose to simply ignore it. (A mathematician would not ignore it. They would, instead, study it!)
The equations you were given are approximate ones. More exact equations can be derived, but the resulting equations would require complex mathematics as they involve solutions containing sine/cosine functions and exponential decay functions. Full solutions for \$T_1\$, \$T_2\$, and \$T_3\$ involve the LambertW function and require applying techniques such as cylindrical decomposition, Cauchy's bound, Richardson's lemma and the Sturm-Richardson method.
These have only recently been developed and published for applications in astrophysics, molecular physics, and condensed matter physics. It's ironic that such a simple combination of electronic parts might require modern developments in mathematics to solve completely. But it does.
Unless you ask for it, I'll avoid adding that mathematical development (I just performed that task last night when going to sleep -- it was definitely a joy for me.) You should instead just accept the approximations you have. They are good enough for most practical purposes.