 Solution

The eigenvalues of $L$ are $\pi /\sqrt{2}$, $-\pi /\sqrt{2}$ and zero, none equal to one. On $E=\mathrm{span}\{1,\cos 2x\}$ the map $L-I$ is an invertible two-dimensional operator; on $E^{\perp }$ it is $-I$. Hence it is invertible on all of $L^2(0,\pi )$. Thus $(L-I)y=\sin x$ has exactly one solution. This is the finite-rank Fredholm alternative; no range obstruction survives the nonresonant identity term.