If $8 \mathrm{cosx}+15\sin \mathrm{x}=15$ and $\cos \mathrm{x}\ne$ then $8 \mathrm{sinx}-15 \cos \mathrm{x}=$

$$\begin{gathered} 8\cos \mathrm{x}+15\sin \mathrm{x}=15 \\ 8\sin \mathrm{x}-15\cos \mathrm{x}=\mathrm{a} \\ {\left(8{\cos }^2 \mathrm{x}+15\sin \mathrm{x}\right)}^2+{\left(8\sin \mathrm{x}-15\cos \mathrm{x}\right)}^2={15}^2+{\mathrm{a}}^2 \\ 64{\cos }^2\mathrm{x}+225{\sin }^2\mathrm{x}+240\sin \mathrm{x} \cos \mathrm{x} \\ +64{\sin }^2\mathrm{x}+225{\cos }^2\mathrm{x}-240\sin \mathrm{x} \cos \mathrm{z}=225+{\mathrm{a}}^2 \\ 64\left({\sin }^2\mathrm{x}+{\cos }^2\right)+225 \left({\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}\right)=225+{\mathrm{a}}^2 \\ 64+225=225+{\mathrm{a}}^2 \\ {\mathrm{a}}^2=64 \Rightarrow \mathrm{a}=-8 \end{gathered}$$