= Solution
Eliminate cis using $[\mathrm{cis}]=C_{\rm tot}-[\mathrm{trans}]$:
$$
\frac{\mathrm d[\mathrm{trans}]}{\mathrm dt}
=k'_1C_{\rm tot}-(k'_1+k'_{-1})[\mathrm{trans}].
$$
Thus
$$
\boxed{c_1=k'_1=1.04\times10^{-3}\,\mathrm{s^{-1}},\qquad c_2=k'_1+k'_{-1}=1.4902\times10^{-3}\,\mathrm{s^{-1}}.}
$$
Both directions restore the perturbed distribution: an excess of trans both increases its reverse consumption and reduces cis available for forward formation. This explains why the <chemical relaxation time> is $1/(k'_1+k'_{-1})$, rather than the reciprocal of only one <rate constant>.
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