Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$.
$\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =$ $\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}$. 
The period of the [tangent](https://artofproblemsolving.com/wiki/index.php/Trigonometry#Tangent) function is $180^\circ$, and the tangent function is [one-to-one](https://artofproblemsolving.com/wiki/index.php/One-to-one) over each period of its domain. 
Thus, $19x \equiv 141 \pmod{180}$. 
Since $19^2 \equiv 361 \equiv 1 \pmod{180}$, multiplying both sides by $19$ yields $x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}$. 
Therefore, the smallest positive solution is $x = \boxed{159}$.