tg2x=>1 inequation

Given the inequality:
$$\tan{\left(2 x \right)} \geq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\tan{\left(2 x \right)} = 1$$
Solve:
Given the equation
$$\tan{\left(2 x \right)} = 1$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{atan}{\left(1 \right)}$$
Or
$$2 x = \pi n + \frac{\pi}{4}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{8}$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{8}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{8}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{\pi}{8}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{8}$$
substitute to the expression
$$\tan{\left(2 x \right)} \geq 1$$
$$\tan{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{8}\right) \right)} \geq 1$$

   /  1   pi       \     
tan|- - + -- + pi*n| >= 1
   \  5   4        /     

but

   /  1   pi       \    
tan|- - + -- + pi*n| < 1
   \  5   4        /    

Then
$$x \leq \frac{\pi n}{2} + \frac{\pi}{8}$$
no execute
the solution of our inequality is:
$$x \geq \frac{\pi n}{2} + \frac{\pi}{8}$$

         _____  
        /
-------•-------
       x1