Electrical – How to solve this diode circuit

It's just the basic operating principle of a diode. An ideal diode would allow current to flow in one direction but block current flow in the other direction. This is based on how it's made, with a p-type region, an n-type region, and a depletion zone in between. Like the bottom diode in this picture:

When you apply a some voltage, in your case 0.9V then the p-type holes and the n-type electrons move into the depletion region because they are repelled by their respective battery terminal. With enough voltage (0.9V in your case) the free electrons in the depletion region get moving and current begins to flow like this:

Now in the ideal case if you were to reverse that battery the opposite will happen and you'll get no current flowing:

In the real world though you can only apply so much reverse voltage or push before you hit the breakdown voltage and current begins to flow freely in the reverse direction. Zener diodes take advantage of this fact and are constructed to breakdown at lower voltages such as your 3.3V.

Sources:
You can read more about how zeners are made here Or see the article I got all the pictures from here

that until input waveform goes above zero at 2π and diode becomes forward biased again.

If the capacitor is charged with Vm the diode does not become forward biased again until almost (3/2)π, to replace the charge that has leaked off through the resistor. If the RC time constant is much longer than the period then that will be close to (3/2)π and the diode only conducts briefly at the positive peaks after initially charging.

Note that this is assuming an ideal diode - a real diode has a significant voltage drop relative to your 2V peak signal.