sin(x/4)<-sqrt3/2 inequation

Given the inequality:
$$\sin{\left(\frac{x}{4} \right)} < \frac{\left(-1\right) \sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(\frac{x}{4} \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
Solve:
Given the equation
$$\sin{\left(\frac{x}{4} \right)} = \frac{\left(-1\right) \sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{4} = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)}$$
$$\frac{x}{4} = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$\frac{x}{4} = 2 \pi n - \frac{\pi}{3}$$
$$\frac{x}{4} = 2 \pi n + \frac{4 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{4}$$
$$x_{1} = 8 \pi n - \frac{4 \pi}{3}$$
$$x_{2} = 8 \pi n + \frac{16 \pi}{3}$$
$$x_{1} = 8 \pi n - \frac{4 \pi}{3}$$
$$x_{2} = 8 \pi n + \frac{16 \pi}{3}$$
This roots
$$x_{1} = 8 \pi n - \frac{4 \pi}{3}$$
$$x_{2} = 8 \pi n + \frac{16 \pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(8 \pi n - \frac{4 \pi}{3}\right) + - \frac{1}{10}$$
=
$$8 \pi n - \frac{4 \pi}{3} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(\frac{x}{4} \right)} < \frac{\left(-1\right) \sqrt{3}}{2}$$
$$\sin{\left(\frac{8 \pi n - \frac{4 \pi}{3} - \frac{1}{10}}{4} \right)} < \frac{\left(-1\right) \sqrt{3}}{2}$$

                            ___ 
    /1    pi         \   -\/ 3  
-sin|-- + -- - 2*pi*n| < -------
    \40   3          /      2   
                         

one of the solutions of our inequality is:
$$x < 8 \pi n - \frac{4 \pi}{3}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < 8 \pi n - \frac{4 \pi}{3}$$
$$x > 8 \pi n + \frac{16 \pi}{3}$$