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