tgx>sqr(3) inequation

Given the inequality:
$$\tan{\left(x \right)} > 9$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(x \right)} = 9$$
Solve:
Given the equation
$$\tan{\left(x \right)} = 9$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{atan}{\left(9 \right)}$$
Or
$$x = \pi n + \operatorname{atan}{\left(9 \right)}$$
, where n - is a integer
$$x_{1} = \pi n + \operatorname{atan}{\left(9 \right)}$$
$$x_{1} = \pi n + \operatorname{atan}{\left(9 \right)}$$
This roots
$$x_{1} = \pi n + \operatorname{atan}{\left(9 \right)}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\pi n + \operatorname{atan}{\left(9 \right)}\right) + - \frac{1}{10}$$
=
$$\pi n - \frac{1}{10} + \operatorname{atan}{\left(9 \right)}$$
substitute to the expression
$$\tan{\left(x \right)} > 9$$
$$\tan{\left(\pi n - \frac{1}{10} + \operatorname{atan}{\left(9 \right)} \right)} > 9$$

tan(-1/10 + pi*n + atan(9)) > 9

Then
$$x < \pi n + \operatorname{atan}{\left(9 \right)}$$
no execute
the solution of our inequality is:
$$x > \pi n + \operatorname{atan}{\left(9 \right)}$$

         _____  
        /
-------ο-------
       x1