Past exam of the differential equations and applications course of the University of Oxford / 2009 / A / 2 / B / c / ii

Here ${\sin }^{2}x=(1-\cos 2x)/2$ belongs to the image. Matching the two coefficients using $L(A+B\cos 2x)=B\pi /2+A\pi \cos 2x$ gives $A=-1/(2\pi )$, $B=1/\pi$. ConsequentlyThere are infinitely many solutions, since any element of the infinite-dimensional kernel may be added.