Find the exact form (Pre-Calculus)

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