cos(x/3)>√2/2 inequation

Given the inequality:
$$\cos{\left(\frac{x}{3} \right)} > \frac{\sqrt{2}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\cos{\left(\frac{x}{3} \right)} = \frac{\sqrt{2}}{2}$$
Solve:
Given the equation
$$\cos{\left(\frac{x}{3} \right)} = \frac{\sqrt{2}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{3} = \pi n + \operatorname{acos}{\left(\frac{\sqrt{2}}{2} \right)}$$
$$\frac{x}{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{\sqrt{2}}{2} \right)}$$
Or
$$\frac{x}{3} = \pi n + \frac{\pi}{4}$$
$$\frac{x}{3} = \pi n - \frac{3 \pi}{4}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{3}$$
$$x_{1} = 3 \pi n + \frac{3 \pi}{4}$$
$$x_{2} = 3 \pi n - \frac{9 \pi}{4}$$
$$x_{1} = 3 \pi n + \frac{3 \pi}{4}$$
$$x_{2} = 3 \pi n - \frac{9 \pi}{4}$$
This roots
$$x_{1} = 3 \pi n + \frac{3 \pi}{4}$$
$$x_{2} = 3 \pi n - \frac{9 \pi}{4}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(3 \pi n + \frac{3 \pi}{4}\right) + - \frac{1}{10}$$
=
$$3 \pi n - \frac{1}{10} + \frac{3 \pi}{4}$$
substitute to the expression
$$\cos{\left(\frac{x}{3} \right)} > \frac{\sqrt{2}}{2}$$
$$\cos{\left(\frac{3 \pi n - \frac{1}{10} + \frac{3 \pi}{4}}{3} \right)} > \frac{\sqrt{2}}{2}$$

                          ___
   /  1    pi       \   \/ 2 
cos|- -- + -- + pi*n| > -----
   \  30   4        /     2  
                        

Then
$$x < 3 \pi n + \frac{3 \pi}{4}$$
no execute
one of the solutions of our inequality is:
$$x > 3 \pi n + \frac{3 \pi}{4} \wedge x < 3 \pi n - \frac{9 \pi}{4}$$

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        /     \  
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       x1      x2