If $\sin(x) =3 \cos(x)$ then what is $\sin(x) \cdot \cos(x)$?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{5}\qquad \textbf{(C)}\ \frac{2}{9}\qquad \textbf{(D)}\ \frac{1}{4}\qquad \textbf{(E)}\ \frac{3}{10}$
In the problem we are given that $\sin{(x)}=3\cos{(x)}$, and we want to find $\sin{(x)}\cos{(x)}$. We can divide both sides of the original equation by $\cos{(x)}$ to get \[\frac{\sin{(x)}}{\cos{(x)}}=\tan{(x)}=3.\]
We can now use right triangle trigonometry to finish the problem.

(Note that this assumes that $x$ is acute. If $x$ is obtuse, then $\sin{(x)}$ is positive and $\cos{(x)}$ is negative, so the equation cannot be satisfied. If $x$ is reflex, then both $\sin{(x)}$ and $\cos{(x)}$ are negative, so the equation is satisfied, but when we find $\sin{(x)}\cos{(x)}$, the two negatives will cancel out and give the same (positive) answer as in the acute case.)
Since the problem asks us to find $\sin{(x)}\cos{(x)}$.
\[\sin{(x)}\cos{(x)}=\left(\frac{3}{\sqrt{10}}\right)\left(\frac{1}{\sqrt{10}}\right)=\frac{3}{10}.\]
So $\boxed{\textbf{(E)}\ \frac{3}{10}}$ is our answer.