 Solution

Rewrite the equation as $(L-I)y=\cos x$. On $E=\mathrm{span}\{\cos 2x,\sin 2x\}$, the operator $L-I$ has eigenvalues $\pi /2-1$ and $\pi /4-1$, both nonzero. On $E^{\perp }$ it equals $-I$. The orthogonal decomposition $L^2(0,\pi )=E\oplus E^{\perp }$ therefore makes $L-I$ bijective with bounded inverse. In particular there is exactly one solution for this forcing. This is also the finite-rank Fredholm alternative: $1$ is not an eigenvalue of $L$, so adding the identity term removes the image obstruction present in $Ly=f$.