sin(3*x)>-sqrt(3)/2 inequation

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

                            ___ 
    /3    pi         \   -\/ 3  
-sin|-- + -- - 2*pi*n| > -------
    \10   3          /      2   
                         

Then
$$x < \frac{2 \pi n}{3} - \frac{\pi}{9}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{2 \pi n}{3} - \frac{\pi}{9} \wedge x < \frac{2 \pi n}{3} + \frac{4 \pi}{9}$$

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