A regular polygon of $n$ sides is inscribed in a circle of radius $R$. The area of the polygon is $3R^2$. Then $n$ equals: 
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 15\qquad \textbf{(E)}\ 18$
Note that the distance from the center of the [circle](https://artofproblemsolving.com/wiki/index.php/Circle) to each of the vertices of the inscribed [regular polygon](https://artofproblemsolving.com/wiki/index.php/Regular_polygon) equals the radius $R$.  Since each side of a regular polygon is the same length, all the angles between the two lines from the center to the two vertices of a side is the same.
That means each of these angles between the two lines from the center to the two vertices of a side equals $\frac{360}{n}$ degrees.  Thus, the area of the polygon is
\[n \cdot \frac{1}{2}R^2\sin\left(\frac{360}{n}^{\circ}\right) = 3R^2\]
Dividing both sides by $R^2$ yields
\[\frac{n}{2}\sin\left(\frac{360}{n}^{\circ}\right) = 3\]
Multiply both sides by $\frac{2}{n}$ to get
\[\sin\left(\frac{360}{n}^{\circ}\right) = \frac{6}{n}\]
At this point, use trial-and-error for each of the answer choices.  When checking $n = 12$, the equation results in $\sin(30^{\circ}) = \frac{1}{2}$, which is correct.  Thus, the answer is $\boxed{\textbf{(C)}}$.