Since $0<x<\pi/2$, both $\sin x$ and $\cos x$ are positive numbers. We have

\[\cot (x)=\frac{\cos x}{\sin x}=\frac{16}{7} \Rightarrow \cos x= \frac{16}{7}\sin x . (1)\]

On the other hand

\[\sin ^2 x +\cos^2 x=1 \Rightarrow \sin^2x+ (\frac{16}{7}\sin x )^2=1\]

\[\Rightarrow \sin^2x+\frac{256}{49}\sin^2x=1 \Rightarrow (1+\frac{256}{49})\sin^2x =1\]

\[\frac{305}{49}\sin^2 x=1 \Rightarrow \sin^2 x=\frac{49}{305} \Rightarrow \sin x= \frac{7}{\sqrt{305}}. (2)\]

Using equation (1) we get

\[\cos x=\frac{16}{7}\sin x=\frac{16}{7}\frac{7}{\sqrt{305}}=\frac{16}{\sqrt{305}}.\]

Now we have

\[\sin 2x =2\sin x \cos x=2\frac{7}{\sqrt{305}} \frac{16}{\sqrt{305}}=\frac{224}{(\sqrt{305})^2}=\frac{224}{305}.\]