1. We are asked to find the smallest positive integer $x$ such that $$\tan(x - 160) = \frac{\cos 50}{1 - \sin 50}.$$ 2. Notice the right side: it resembles the tangent half-angle identity $$\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$$ or variants. Let's simplify the right side using known identities. 3. Using the identity $$\tan \alpha = \frac{\sin \alpha}{\cos \alpha},$$ rewrite the denominator: $$1 - \sin 50 = \frac{(1 - \sin 50)(1 + \sin 50)}{1 + \sin 50} = \frac{1 - \sin^2 50}{1 + \sin 50} = \frac{\cos^2 50}{1 + \sin 50}.$$ 4. So $$\frac{\cos 50}{1 - \sin 50} = \frac{\cos 50}{\frac{\cos^2 50}{1 + \sin 50}} = \frac{\cos 50 (1 + \sin 50)}{\cos^2 50} = \frac{1 + \sin 50}{\cos 50}.$$ 5. Thus $$\tan(x - 160) = \frac{1 + \sin 50}{\cos 50}$$ which can be recognized as the tangent of an angle as well. 6. Using the tangent addition formula: $$\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}.$$ But more simply, recall the identity for tangent of sum: $$\tan(45^\circ + \theta) = \frac{1 + \tan \theta}{1 - \tan \theta}.$$ Alternatively, observe that $$\frac{1 + \sin 50}{\cos 50} = \frac{\cos 40 + \sin 40 + 1}{\cos 50}$$ is not straightforward, so the best approach is to find angle $y$ such that $$\tan y = \frac{1 + \sin 50}{\cos 50}.$$ 7. Calculate numerically: $\sin 50 \approx 0.7660,$ $\cos 50 \approx 0.6428,$ so $$\frac{1 + 0.7660}{0.6428} \approx \frac{1.7660}{0.6428} \approx 2.748.$$ 8. Find $y = \arctan(2.748) \approx 70.016^\circ.$ 9. Recall from the problem: $$\tan(x - 160) = \tan y$$ implying $$x - 160 = y + k \times 180,\quad k \in \mathbb{Z}.$$ 10. We want the smallest positive integer $x$, so pick $k=0$: $$x = 160 + y \approx 160 + 70.016 = 230.016.$$ Since we want an integer, try $x = 230.$ 11. If rounded down is not positive (but it is >0), try $k=-1$ to check smaller solutions: $$x = 160 + y - 180 = -19.984,$$ negative, discard. 12. Test $x=230:$ it is positive integer and satisfies approximately. Thus, the smallest positive integer solution is $$\boxed{230}.$$