sin7x>√3/2 inequation

Given the inequality:
$$\sin{\left(7 x \right)} > \frac{\sqrt{3}}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin{\left(7 x \right)} = \frac{\sqrt{3}}{2}$$
Solve:
Given the equation
$$\sin{\left(7 x \right)} = \frac{\sqrt{3}}{2}$$
- this is the simplest trigonometric equation
This equation is transformed to
$$7 x = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)}$$
$$7 x = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{3}}{2} \right)} + \pi$$
Or
$$7 x = 2 \pi n + \frac{\pi}{3}$$
$$7 x = 2 \pi n + \frac{2 \pi}{3}$$
, where n - is a integer
Divide both parts of the equation by
$$7$$
$$x_{1} = \frac{2 \pi n}{7} + \frac{\pi}{21}$$
$$x_{2} = \frac{2 \pi n}{7} + \frac{2 \pi}{21}$$
$$x_{1} = \frac{2 \pi n}{7} + \frac{\pi}{21}$$
$$x_{2} = \frac{2 \pi n}{7} + \frac{2 \pi}{21}$$
This roots
$$x_{1} = \frac{2 \pi n}{7} + \frac{\pi}{21}$$
$$x_{2} = \frac{2 \pi n}{7} + \frac{2 \pi}{21}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$\left(\frac{2 \pi n}{7} + \frac{\pi}{21}\right) + - \frac{1}{10}$$
=
$$\frac{2 \pi n}{7} - \frac{1}{10} + \frac{\pi}{21}$$
substitute to the expression
$$\sin{\left(7 x \right)} > \frac{\sqrt{3}}{2}$$
$$\sin{\left(7 \left(\frac{2 \pi n}{7} - \frac{1}{10} + \frac{\pi}{21}\right) \right)} > \frac{\sqrt{3}}{2}$$

                            ___
   /  7    pi         \   \/ 3 
sin|- -- + -- + 2*pi*n| > -----
   \  10   3          /     2  
                          

Then
$$x < \frac{2 \pi n}{7} + \frac{\pi}{21}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{2 \pi n}{7} + \frac{\pi}{21} \wedge x < \frac{2 \pi n}{7} + \frac{2 \pi}{21}$$

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       x1      x2