= Solution
Rewrite the equation as $(L-I)y=\cos x$. On $E=\operatorname{span}\{\cos2x,\sin2x\}$, 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 \b[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 $Ly=f$.
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