tgx<√3

To solve the inequality \( \tan(x) < \sqrt{3} \), we need to determine the values of \( x \) for which the tangent function is less than \( \sqrt{3} \).

First, we recognize that: \[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \] This means we need to find the ranges of \( x \) where \( \tan(x) \) is less than \( \sqrt{3} \).

The tangent function is periodic with a period of \( \pi \). Therefore, the intervals in which \( \tan(x) < \sqrt{3} \) can be evaluated based on the behavior of the tangent function in one period.

1. First period (from \( 0 \) to \( \pi \)):
   • \( \tan(x) < \sqrt{3} \) for \( 0 < x < \frac{\pi}{3} \)
   • \( \tan(x) > \sqrt{3} \) for \( \frac{\pi}{3} < x < \frac{\pi}{2} \)

Since the tangent function has vertical asymptotes at \( x = \frac{\pi}{2} + k\pi \) (where \( k \) is any integer), we need to consider these intervals in every period:

2. General solution:
   • For each interval from \( n\pi \) to \( (n+1)\pi \): \[ x \in \left( n\pi, n\pi + \frac{\pi}{3} \right) \quad \text{for } n \in \mathbb{Z} \]
   • This contains all solutions where \( \tan(x) < \sqrt{3} \).

So the complete solution set is: \[ x \in \bigcup_{n \in \mathbb{Z}} \left( n\pi, n\pi + \frac{\pi}{3} \right) \]

To summarize, the values of \( x \) for which \( \tan(x) < \sqrt{3} \) are within the intervals given above.