Answer
$(D^2+4)(D^2+16)$ is the annihilator of $F(x)$.
Work Step by Step
We have: $F(x)=\sin x \cos^3 x $
Re-write as: $F(x)=\sin x (\dfrac{3\cos x-\cos 3x}{4})=\dfrac{1}{2}\sin 2x-\dfrac{\sin 4x}{8}$
So, $D^2+4$ is the annihilator of $\dfrac{\sin 2x}{2}$ and $D^2+16$ is the annihilator of $\dfrac{-\sin 4x}{8}$ .
Therefore, $(D^2+4)(D^2+16)$ is the annihilator of $F(x)$.
