cosx(tg2x-1)≤0 inequation

Given the inequality:
$$\left(\tan{\left(2 x \right)} - 1\right) \cos{\left(x \right)} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(\tan{\left(2 x \right)} - 1\right) \cos{\left(x \right)} = 0$$
Solve:
$$x_{1} = - \frac{\pi}{2}$$
$$x_{2} = \frac{\pi}{8}$$
$$x_{3} = \frac{\pi}{2}$$
$$x_{1} = - \frac{\pi}{2}$$
$$x_{2} = \frac{\pi}{8}$$
$$x_{3} = \frac{\pi}{2}$$
This roots
$$x_{1} = - \frac{\pi}{2}$$
$$x_{2} = \frac{\pi}{8}$$
$$x_{3} = \frac{\pi}{2}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{2} - \frac{1}{10}$$
=
$$- \frac{\pi}{2} - \frac{1}{10}$$
substitute to the expression
$$\left(\tan{\left(2 x \right)} - 1\right) \cos{\left(x \right)} \leq 0$$
$$\left(-1 + \tan{\left(2 \left(- \frac{\pi}{2} - \frac{1}{10}\right) \right)}\right) \cos{\left(- \frac{\pi}{2} - \frac{1}{10} \right)} \leq 0$$

-(-1 - tan(1/5))*sin(1/10) <= 0

but

-(-1 - tan(1/5))*sin(1/10) >= 0

Then
$$x \leq - \frac{\pi}{2}$$
no execute
one of the solutions of our inequality is:
$$x \geq - \frac{\pi}{2} \wedge x \leq \frac{\pi}{8}$$

         _____           _____  
        /     \         /
-------•-------•-------•-------
       x1      x2      x3

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq - \frac{\pi}{2} \wedge x \leq \frac{\pi}{8}$$
$$x \geq \frac{\pi}{2}$$