Question

Tgx>-√3 срочно !!!!!!!!

Answer 1

$tgx>-\sqrt{3}\\\\arctg(-\sqrt{3})+\pi n$

x ∈$(-\frac{\pi }{3}+\pi n;\frac{\pi }{2}+\pi n),nez$  

Answer 2

It seems like you've entered a mathematical expression: $Tgx > -\sqrt{3}$. "Tgx" is likely a typo, as the standard abbreviation for the tangent function is usually written as "tan" or "tg" (depending on the language).

Assuming you meant $\tan(x) > -\sqrt{3}$, this is an inequality involving the tangent function. To solve it, you need to find the values of $x$ that satisfy the inequality.

The tangent function is negative in the second and fourth quadrants, and its absolute value is always less than or equal to 1. Therefore, $\tan(x) > -\sqrt{3}$ is satisfied for all values of $x$ in the second and fourth quadrants.

In mathematical notation, the solution can be expressed as:

$x \in \left( \left(2n - \frac{1}{2}\right) \pi, 2n\pi \right)$ where $n$ is an integer.

If you have a specific value for $x$ and want to evaluate whether it satisfies the inequality, please provide more details.