cos(6x-1)>0.5 inequation

Given the inequality:
$$\cos{\left(6 x - 1 \right)} > \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(6 x - 1 \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\cos{\left(6 x - 1 \right)} = \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$6 x - 1 = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$6 x - 1 = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
Or
$$6 x - 1 = \pi n + \frac{\pi}{3}$$
$$6 x - 1 = \pi n - \frac{2 \pi}{3}$$
, where n - is a integer
Move
$$-1$$
to right part of the equation
with the opposite sign, in total:
$$6 x = \pi n + 1 + \frac{\pi}{3}$$
$$6 x = \pi n - \frac{2 \pi}{3} + 1$$
Divide both parts of the equation by
$$6$$
$$x_{1} = \frac{\pi n}{6} + \frac{1}{6} + \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{9} + \frac{1}{6}$$
$$x_{1} = \frac{\pi n}{6} + \frac{1}{6} + \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{9} + \frac{1}{6}$$
This roots
$$x_{1} = \frac{\pi n}{6} + \frac{1}{6} + \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{6} - \frac{\pi}{9} + \frac{1}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{6} + \frac{1}{6} + \frac{\pi}{18}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{6} + \frac{1}{15} + \frac{\pi}{18}$$
substitute to the expression
$$\cos{\left(6 x - 1 \right)} > \frac{1}{2}$$
$$\cos{\left(6 \left(\frac{\pi n}{6} + \frac{1}{15} + \frac{\pi}{18}\right) - 1 \right)} > \frac{1}{2}$$

   /  3   pi       \      
cos|- - + -- + pi*n| > 1/2
   \  5   3        /      

Then
$$x < \frac{\pi n}{6} + \frac{1}{6} + \frac{\pi}{18}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{\pi n}{6} + \frac{1}{6} + \frac{\pi}{18} \wedge x < \frac{\pi n}{6} - \frac{\pi}{9} + \frac{1}{6}$$

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       x1      x2