sin(4x+2)<-sqrt(2)/2 inequation

Given the inequality:
$$\sin{\left(4 x + 2 \right)} < \frac{\left(-1\right) \sqrt{2}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(4 x + 2 \right)} = \frac{\left(-1\right) \sqrt{2}}{2}$$
Solve:
Given the equation
$$\sin{\left(4 x + 2 \right)} = \frac{\left(-1\right) \sqrt{2}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$4 x + 2 = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{2}}{2} \right)}$$
$$4 x + 2 = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{2}}{2} \right)} + \pi$$
Or
$$4 x + 2 = 2 \pi n - \frac{\pi}{4}$$
$$4 x + 2 = 2 \pi n + \frac{5 \pi}{4}$$
, where n - is a integer
Move
$$2$$
to right part of the equation
with the opposite sign, in total:
$$4 x = 2 \pi n - 2 - \frac{\pi}{4}$$
$$4 x = 2 \pi n - 2 + \frac{5 \pi}{4}$$
Divide both parts of the equation by
$$4$$
$$x_{1} = \frac{\pi n}{2} - \frac{1}{2} - \frac{\pi}{16}$$
$$x_{2} = \frac{\pi n}{2} - \frac{1}{2} + \frac{5 \pi}{16}$$
$$x_{1} = \frac{\pi n}{2} - \frac{1}{2} - \frac{\pi}{16}$$
$$x_{2} = \frac{\pi n}{2} - \frac{1}{2} + \frac{5 \pi}{16}$$
This roots
$$x_{1} = \frac{\pi n}{2} - \frac{1}{2} - \frac{\pi}{16}$$
$$x_{2} = \frac{\pi n}{2} - \frac{1}{2} + \frac{5 \pi}{16}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{\pi n}{2} - \frac{1}{2} - \frac{\pi}{16}\right) + - \frac{1}{10}$$
=
$$\frac{\pi n}{2} - \frac{3}{5} - \frac{\pi}{16}$$
substitute to the expression
$$\sin{\left(4 x + 2 \right)} < \frac{\left(-1\right) \sqrt{2}}{2}$$
$$\sin{\left(4 \left(\frac{\pi n}{2} - \frac{3}{5} - \frac{\pi}{16}\right) + 2 \right)} < \frac{\left(-1\right) \sqrt{2}}{2}$$

                           ___ 
    /2   pi         \   -\/ 2  
-sin|- + -- - 2*pi*n| < -------
    \5   4          /      2   
                        

one of the solutions of our inequality is:
$$x < \frac{\pi n}{2} - \frac{1}{2} - \frac{\pi}{16}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi n}{2} - \frac{1}{2} - \frac{\pi}{16}$$
$$x > \frac{\pi n}{2} - \frac{1}{2} + \frac{5 \pi}{16}$$