2*sin(6*x)+sqrt(3)<0 inequation

Given the inequality:
$$2 \sin{\left(6 x \right)} + \sqrt{3} < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \sin{\left(6 x \right)} + \sqrt{3} = 0$$
Solve:
Given the equation
$$2 \sin{\left(6 x \right)} + \sqrt{3} = 0$$
- this is the simplest trigonometric equation

Move sqrt(3) to right part of the equation

with the change of sign in sqrt(3)

We get:
$$2 \sin{\left(6 x \right)} = - \sqrt{3}$$

Divide both parts of the equation by 2

The equation is transformed to
$$\sin{\left(6 x \right)} = - \frac{\sqrt{3}}{2}$$
This equation is transformed to
$$6 x = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$6 x = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$6 x = 2 \pi n - \frac{\pi}{3}$$
$$6 x = 2 \pi n + \frac{4 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$6$$
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
This roots
$$x_{1} = \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x_{2} = \frac{\pi n}{3} + \frac{2 \pi}{9}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{3} - \frac{\pi}{18}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{3} - \frac{\pi}{18} - \frac{1}{10}$$
substitute to the expression
$$2 \sin{\left(6 x \right)} + \sqrt{3} < 0$$
$$2 \sin{\left(6 \left(\frac{\pi n}{3} - \frac{\pi}{18} - \frac{1}{10}\right) \right)} + \sqrt{3} < 0$$

  ___        /3   pi         \    
\/ 3  - 2*sin|- + -- - 2*pi*n| < 0
             \5   3          /    

one of the solutions of our inequality is:
$$x < \frac{\pi n}{3} - \frac{\pi}{18}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{3} - \frac{\pi}{18}$$
$$x > \frac{\pi n}{3} + \frac{2 \pi}{9}$$