cos2x<=-1/2 inequation

Given the inequality:
$$\cos{\left(2 x \right)} \leq - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(2 x \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(2 x \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$2 x = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
Or
$$2 x = \pi n + \frac{2 \pi}{3}$$
$$2 x = \pi n - \frac{\pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{3}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{6}$$
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{3}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{6}$$
This roots
$$x_{1} = \frac{\pi n}{2} + \frac{\pi}{3}$$
$$x_{2} = \frac{\pi n}{2} - \frac{\pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} + \frac{\pi}{3}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{3}$$
substitute to the expression
$$\cos{\left(2 x \right)} \leq - \frac{1}{2}$$
$$\cos{\left(2 \left(\frac{\pi n}{2} - \frac{1}{10} + \frac{\pi}{3}\right) \right)} \leq - \frac{1}{2}$$

    /  1   pi       \        
-sin|- - + -- + pi*n| <= -1/2
    \  5   6        /        

but

    /  1   pi       \        
-sin|- - + -- + pi*n| >= -1/2
    \  5   6        /        

Then
$$x \leq \frac{\pi n}{2} + \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$x \geq \frac{\pi n}{2} + \frac{\pi}{3} \wedge x \leq \frac{\pi n}{2} - \frac{\pi}{6}$$

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       x1      x2