\(\left(16-x^2\right)\sqrt{x-3}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\16-x^2\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x\in(-\infty;-4]\cup[4;+\infty)\end{matrix}\right.\)
\(\Leftrightarrow\left\{3\right\}\cup[4;+\infty)\)