Both domains describe the same thing, but seen from different perspectives. I like to think of the frequency domain as the time domain skewed through 90°.
In the time domain you have your typical signal waveform - the X axis is time passing, and the Y axis is the level of the signal at any one point in time.
Now, a signal is made up of lots of frequencies. Take, for example, a "simple" square wave. As time passes it alternates between a low and a high level. Very simple. However, fundamentally, there is only the sine wave, so that square wave must be made up of multiple different sine waves added together.
The frequency domain is a snapshot of a brief moment in time, and describes what frequencies are present in the signal during that time. A square wave may be fundamentally 1KHz, but in the frequency domain we can see that it is made up of sine waves at 1KHz (the fundamental frequency), 3KHz, 5KHz, 7KHz, etc - each one smaller than the other.
Another good way of looking at the relationship between the two domains is to think about frequency and period. In the frequency domain a sine wave may be 1KHz, but in the time domain it has a period of 1ms. They both describe the same thing, but looked at from different directions.
I would be very grateful if someone could give a concise answer, how
to solve it, and if it is not possible, why it is not possible.
It is possible. For \$t \lt 0\$, the switch is in position 1 and you have a series RC circuit in sinusoidal steady state.
The phasor voltage across the capacitor in a series RC circuit is given by:
$$V_c(\omega) = V_s\frac{1}{1 + j\omega RC} = V_s \frac{e^{i\phi}}{\sqrt{1 + (\omega RC)^2}}$$
where \$\tan \phi = -\omega RC\$.
In the case of the circuit in your question
$$V_s = 10\sqrt{2}\;e^{i\frac{\pi}{4}}$$
$$\omega RC = 1$$
Before the switch changes state, the RC circuit is in steady state and so
$$V_c = 10\sqrt{2}\;e^{i\frac{\pi}{4}}\frac{e^{-i\frac{\pi}{4}}}{\sqrt{1 + 1}} = 10\mathrm V$$
Converting to the time domain, the capacitor voltage is then
$$v_C(t) = 10\mathrm V \cos(1000\cdot t)\;, t \le 0$$
and thus the capacitor voltage, just before the switch changes state is
$$v_C(0-) = 10 \cos(0-) = 10 \mathrm V $$
Since the voltage across a capacitor cannot instantaneously change (for finite current through), the voltage across the capacitor just after the switch changes state is unchanged.
$$v_C(0+) = v_C(0-) = 10\mathrm V $$
This is the initial capacitor voltage that is required to find the solution for \$t \gt 0\$
Best Answer
The time domain is pretty easy. \$ I_0 sin(wt) \$ The signal varies with time \$t\$ with the frequency \$w\$. The problem with this notation is that i's difficult to deal with delays and comparing the original signal to the delayed one \$ I_0 sin(wt - T) \$ for example. The solution is to transform the signal to the phasor domain. This gives the ability to compare signal timing in degrees instead of time. For example two sine waves that are 360 degrees out of phase line up again. The downside of straight phasors is that every signal in the analysis must be at the same frequency and there is no compensation for components that are frequency sensitive (inductors and capacitors).
The frequency domain transforms all frequency sensitive components to compensate them for frequency. The frequency domain still uses phasor notation. If frequency domain you can also deal with multiple frequency signal but to do that you must use super position.
To move from time to frequency just use a Laplace transform. To move from time to phasor figure out how much delay or advance the signal has \$ degress/360 = T/period \$