sinx>-0.5 inequation

Given the inequality:
$$\sin{\left(x \right)} > - \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(x \right)} = - \frac{1}{2}$$
Solve:
Given the equation
$$\sin{\left(x \right)} = - \frac{1}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = 2 \pi n + \operatorname{asin}{\left(- \frac{1}{2} \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(- \frac{1}{2} \right)} + \pi$$
Or
$$x = 2 \pi n - \frac{\pi}{6}$$
$$x = 2 \pi n + \frac{7 \pi}{6}$$
, where n - is a integer
$$x_{1} = 2 \pi n - \frac{\pi}{6}$$
$$x_{2} = 2 \pi n + \frac{7 \pi}{6}$$
$$x_{1} = 2 \pi n - \frac{\pi}{6}$$
$$x_{2} = 2 \pi n + \frac{7 \pi}{6}$$
This roots
$$x_{1} = 2 \pi n - \frac{\pi}{6}$$
$$x_{2} = 2 \pi n + \frac{7 \pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(2 \pi n - \frac{\pi}{6}\right) + - \frac{1}{10}$$
=
$$2 \pi n - \frac{\pi}{6} - \frac{1}{10}$$
substitute to the expression
$$\sin{\left(x \right)} > - \frac{1}{2}$$
$$\sin{\left(2 \pi n - \frac{\pi}{6} - \frac{1}{10} \right)} > - \frac{1}{2}$$

    /1    pi         \       
-sin|-- + -- - 2*pi*n| > -1/2
    \10   6          /       

Then
$$x < 2 \pi n - \frac{\pi}{6}$$
no execute
one of the solutions of our inequality is:
$$x > 2 \pi n - \frac{\pi}{6} \wedge x < 2 \pi n + \frac{7 \pi}{6}$$

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       x1      x2