sin(x/6)>sqrt(3)/2 inequation

Given the inequality:
$$\sin{\left(\frac{x}{6} \right)} > \frac{\sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{6} \right)} = \frac{\sqrt{3}}{2}$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{6} \right)} = \frac{\sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{6} = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)}$$
$$\frac{x}{6} = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$\frac{x}{6} = 2 \pi n + \frac{\pi}{3}$$
$$\frac{x}{6} = 2 \pi n + \frac{2 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{6}$$
$$x_{1} = 12 \pi n + 2 \pi$$
$$x_{2} = 12 \pi n + 4 \pi$$
$$x_{1} = 12 \pi n + 2 \pi$$
$$x_{2} = 12 \pi n + 4 \pi$$
This roots
$$x_{1} = 12 \pi n + 2 \pi$$
$$x_{2} = 12 \pi n + 4 \pi$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(12 \pi n + 2 \pi\right) + - \frac{1}{10}$$
=
$$12 \pi n - \frac{1}{10} + 2 \pi$$
substitute to the expression
$$\sin{\left(\frac{x}{6} \right)} > \frac{\sqrt{3}}{2}$$
$$\sin{\left(\frac{12 \pi n - \frac{1}{10} + 2 \pi}{6} \right)} > \frac{\sqrt{3}}{2}$$

                            ___
   /  1    pi         \   \/ 3 
sin|- -- + -- + 2*pi*n| > -----
   \  60   3          /     2  
                          

Then
$$x < 12 \pi n + 2 \pi$$
no execute
one of the solutions of our inequality is:
$$x > 12 \pi n + 2 \pi \wedge x < 12 \pi n + 4 \pi$$

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        /     \  
-------ο-------ο-------
       x1      x2