If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$, then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$
We divide both sides of the [equation](https://artofproblemsolving.com/wiki/index.php/Equation) by $\cos{2x}\times\cos{3x}$ to get $\frac{\sin{2x}\times\sin{3x}}{\cos{2x}\times\cos{3x}}=1$, or $\tan{2x}\times\tan{3x}=1$.
This looks a lot like the formula relating the slopes of two [perpendicular](https://artofproblemsolving.com/wiki/index.php/Perpendicular) [lines](https://artofproblemsolving.com/wiki/index.php/Line), which is $m_1\times m_2=-1$, where $m_1$ and $m_2$ are the [slopes](https://artofproblemsolving.com/wiki/index.php/Slope). It's made even more relatable by the fact that the [tangent](https://artofproblemsolving.com/wiki/index.php/Tangent) of an [angle](https://artofproblemsolving.com/wiki/index.php/Angle) can be defined by the slope of the line that makes that angle with the [x-axis](https://artofproblemsolving.com/wiki/index.php/X-axis).
We can make this look even more like the slope formula by multiplying both sides by $-1$:
$\tan{2x}\times-\tan{3x}=-1$, and using the [trigonometric identity](https://artofproblemsolving.com/wiki/index.php/Trigonometric_identity) $-\tan{x}=\tan{-x}$, we have $\tan{2x}\times\tan{-3x}=-1$.
Now it's time for a diagram:

Since the product of the two slopes, $\tan{2x}$ and $\tan{-3x}$, is $-1$, the lines are perpendicular, and the [angle](https://artofproblemsolving.com/wiki/index.php/Angle) between them is $\frac{\pi}{2}$. The angle between them is also $2x+3x=5x$, so $5x=\frac{\pi}{2}$ and $x=\frac{\pi}{10}$, or $18^\circ, \boxed{\text{A}}$.
NOTE: To show that $18^\circ$ is not the only solution, we can also set $5x$ equal to another angle measure congruent to $\frac{\pi}{2}$, such as $\frac{5\pi}{2}$, yielding another solution as $\frac{\pi}{2}=90^\circ$, which clearly is a solution to the equation.