sin6x<0 inequation

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

with the change of sign in 0

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

sin(-3/5 + 2*pi*n) < 0

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

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{3}$$
$$x > \frac{\pi n}{3} + \frac{\pi}{6}$$