The minimum value of $(3\sin \mathrm{x}-4\cos \mathrm{x}-10)(3\sin \mathrm{x}+4\cos \mathrm{x}-10)$ is

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=9{\sin }^2\mathrm{x}-16{\cos }^2\mathrm{x}-10(3\sin \mathrm{x}-4\cos \mathrm{x})-10(3\sin \mathrm{x}+4\cos \mathrm{x})+100\\ \\ =25{\sin }^2\mathrm{x}-60\sin \mathrm{x}+84\\ =(5\sin \mathrm{x}-6{)}^2+48\end{array}$

The minimum value of f (x) occurs when sin x = 1 .
Therefore, the minimum value of f (x) is 49.