= Solution
Orient the unit circle counterclockwise and use the standard principal-value Cauchy operator
$$
K\phi(t)=\frac1{\pi i}\operatorname{PV}\int_C\frac{\phi(\tau)}{\tau-t}\,d\tau.
$$
For a Fourier/Laurent mode $t^n$ it gives $+t^n$ when $n\ge0$ and $-t^n$ when $n<0$. This follows from the interior/exterior Cauchy projections, or by direct indentation of the pole. Decompose $\phi=\phi_++\phi_-$ into its nonnegative and negative modes, so $K\phi=\phi_+-\phi_-$. Let
$$
J=\frac1{2\pi i}\int_C(\tau+2\tau^{-1})\phi(\tau)\,d\tau=c_{-2}+2c_0.
$$
The printed equation reduces to
$$
2t\phi_++2t^{-1}\phi_--J(t+t^{-1})=2t^{-1}.
$$
Positive-power coefficients force $\phi_+=J/2$, a constant; powers at most $-2$ force $\phi_-=0$. The $t^{-1}$ coefficient then gives $-J=2$, and the integral definition is consistent with $c_0=-1$. Thus
$$
\boxed{\phi(t)=-1\quad(t\in C).}
$$
Direct substitution has $K\phi=-1$, $J=-2$ and gives $2t^{-1}$ as required. The same projection argument on the homogeneous equation forces $J=0$ and both components zero, proving uniqueness in the usual Hölder or $L^2$ class for this singular-integral problem.
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