The value of n ∈ I for which the function $\mathrm{f}\left(\mathrm{x}\right)=\frac{\sin \mathrm{nx}}{\sin \left(\frac{\mathrm{x}}{\mathrm{n}}\right)} \mathrm{has} 4\mathrm{\pi }$ as its period is

For n = 2, we have
$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{\sin 2\mathrm{x}}{\sin \left(\frac{\mathrm{x}}{2}\right)}=\frac{2\sin \mathrm{xcos}\mathrm{x}}{\sin \left(\frac{\mathrm{x}}{2}\right)} \\ =\frac{4\sin \mathrm{x}/2\cos \mathrm{x}/2\cos \mathrm{x}}{\sin \left(\frac{\mathrm{x}}{2}\right)}=4\cos \frac{\mathrm{x}}{2}\cos \mathrm{x} \end{gathered}$$
The period of cos x is 2$\mathrm{\pi }$ and that of cos $\frac{\mathrm{x}}{2} \mathrm{is} \frac{2\mathrm{\pi }}{1/2}=4\mathrm{\pi }$
Hence, the period of f (x) is 4$\mathrm{\pi }$.