Why don't phasors give the transient state?
Phasor analysis allows us to analyze a circuit's response to a sinusoidal steady-state response at a given single frequency. We represent a time-domain voltage \$V(t) = V_{0}cos(\omega t + \phi)\$ in phasor form by transforming it into a complex exponential via Euler's formula...

$$A\cos(\omega t + \phi) = Re\{Ae^{j( \omega t~+~\phi)}\} = Re\{Ae^{j\phi}e^{j\omega t}\}$$

...and then ignoring the frequency/time dependence (since we assume everything in the circuit is excited by a steady sinusoid of the same frequency). Thus,

$$V= Ae^{j\phi}$$

This \$V\$ is what we call a phasor. We can represent any current or voltage as a phasor. In order to recover a time-domain representation from a phasor, you can multiply it by \$e^{j\omega t}\$ and then take the real part. Note that sometimes for the sake of brevity/familiarity in calculating power, we also convert the amplitude of the phasor into RMS values (divide magnitude by \$\sqrt{2}\$ for a sinusoid). Phasors allow us to use analogous DC analysis techniques to recover transfer functions of linear circuits (by using impedances). Using superposition, we can use Fourier analysis to analyze a circuit's complete steady-state response as a sum of its steady-state response due to different frequency components.

It's useful to note the relation of phasors to the Laplace representation of a circuit. The Laplace representation of a circuit uses the variable \$s = \sigma + j\omega\$. Note that for \$\sigma = 0\$, the transfer function of the Laplace representation of a circuit reduces down to the phasor representation. This is a good indication that the real part of \$s\$ represents a transient response (and this can be easily observed by noting that \$e^{at}\$ for any real \$a\$ will result in a real value that either exponentially grows or decays). Note that the Laplace representation is a more general representation of a circuit that includes both transient and steady-state responses. Likewise, it's good to note that the Fourier transform is just a special case of the more general Laplace transform (the case where \$\sigma = 0\$ in \$s = j\omega\$).