= Solution
Every image of $L$ is a <linear combination> $a\cos2x+b\sin2x$. The function $\cos x$ is not such a combination. Indeed these functions are continuous, so equality almost everywhere would imply equality everywhere; evaluation at the endpoints would require simultaneously $a=1$ and $a=-1$. Therefore the forcing is outside the <image> and \b[there is no solution]. <Orthogonality> to the <kernel> is the equivalent <image> test for this <self-adjoint> <finite-rank operator>.
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