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

The image of $L$ consists only of $a+b\cos 2x$. If $\sin x$ were such a function, its endpoint values would give $a+b=0$ and its midpoint value $a-b=1$, forcing $a=1/2$, $b=-1/2$. That combination is ${\sin }^{2}x$, which differs from $\sin x$, for example at $x=\pi /6$. Continuity makes the same argument valid for equality almost everywhere. Thus the forcing is outside the image and no solution exists.