sin(t)<-sqrt(2)/2 inequation

Given the inequality:
$$\sin{\left(t \right)} < \frac{\left(-1\right) \sqrt{2}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(t \right)} = \frac{\left(-1\right) \sqrt{2}}{2}$$
Solve:
Given the equation
$$\sin{\left(t \right)} = \frac{\left(-1\right) \sqrt{2}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$t = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{2}}{2} \right)}$$
$$t = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{2}}{2} \right)} + \pi$$
Or
$$t = 2 \pi n - \frac{\pi}{4}$$
$$t = 2 \pi n + \frac{5 \pi}{4}$$
, where n - is a integer
$$t_{1} = 2 \pi n - \frac{\pi}{4}$$
$$t_{2} = 2 \pi n + \frac{5 \pi}{4}$$
$$t_{1} = 2 \pi n - \frac{\pi}{4}$$
$$t_{2} = 2 \pi n + \frac{5 \pi}{4}$$
This roots
$$t_{1} = 2 \pi n - \frac{\pi}{4}$$
$$t_{2} = 2 \pi n + \frac{5 \pi}{4}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n - \frac{\pi}{4}\right) + - \frac{1}{10}$$
=
$$2 \pi n - \frac{\pi}{4} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(t \right)} < \frac{\left(-1\right) \sqrt{2}}{2}$$
$$\sin{\left(2 \pi n - \frac{\pi}{4} - \frac{1}{10} \right)} < \frac{\left(-1\right) \sqrt{2}}{2}$$

                            ___ 
    /1    pi         \   -\/ 2  
-sin|-- + -- - 2*pi*n| < -------
    \10   4          /      2   
                         

one of the solutions of our inequality is:
$$t < 2 \pi n - \frac{\pi}{4}$$

 _____           _____          
      \         /
-------ο-------ο-------
       t1      t2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$t < 2 \pi n - \frac{\pi}{4}$$
$$t > 2 \pi n + \frac{5 \pi}{4}$$