= Solution
Differentiating the two equations for a fundamental matrix of solutions gives
$$
\psi_{\rho\tau}=(U_\tau+UV)\psi,\qquad\psi_{\tau\rho}=(V_\rho+VU)\psi.
$$
Equality for all independent initial vectors is equivalent to the <zero-curvature condition>
$$
\boxed{U_\tau-V_\rho+[U,V]=0.}
$$
A single possibly zero solution vector alone would not imply a matrix identity; compatibility is understood for the complete linear system.
Write $q=\phi_\rho$, $s=\sin\phi$, $c=\cos\phi$ and use <Pauli matrices>:
$$
U=i\lambda\sigma_3+\frac{iq}{2}\sigma_1,\qquad V=\frac1{4i\lambda}(c\sigma_3+s\sigma_2),\quad\lambda\ne0.
$$
The matrix products give
$$
[U,V]=-\frac{i s}{2}\sigma_1-\frac{i q c}{4\lambda}\sigma_2+\frac{i q s}{4\lambda}\sigma_3,
$$
whereas $V_\rho=-(iqc/(4\lambda))\sigma_2+(iqs/(4\lambda))\sigma_3$. Thus these last two components cancel, and the remaining curvature is $(i/2)(\phi_{\rho\tau}-\sin\phi)\sigma_1$. Therefore
$$
\boxed{\phi_{\rho\tau}=\sin\phi,}
$$
the <Sine-Gordon equation> in light-cone coordinates.
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