7sin(x/2)^2>1 inequation

Given the inequality:
$$7 \sin^{2}{\left(\frac{x}{2} \right)} > 1$$
To solve this inequality, we must first solve the corresponding equation:
$$7 \sin^{2}{\left(\frac{x}{2} \right)} = 1$$
Solve:
Given the equation
$$7 \sin^{2}{\left(\frac{x}{2} \right)} = 1$$
transform
$$\frac{5}{2} - \frac{7 \cos{\left(x \right)}}{2} = 0$$
$$7 \sin^{2}{\left(\frac{x}{2} \right)} - 1 = 0$$
Do replacement
$$w = \sin{\left(\frac{x}{2} \right)}$$
Given the equation
$$7 \sin^{2}{\left(\frac{x}{2} \right)} - 1 = 0$$
Because equation degree is equal to = 2 - contains the even number 2 in the numerator, then
the equation has two real roots.
Get the root 2-th degree of the equation sides:
We get:
$$\sqrt{7} \sqrt{\left(0 w + \sin{\left(\frac{x}{2} \right)}\right)^{2}} = \sqrt{1}$$
$$\sqrt{7} \sqrt{\left(0 w + \sin{\left(\frac{x}{2} \right)}\right)^{2}} = \left(-1\right) \sqrt{1}$$
or
$$\sqrt{7} \sin{\left(\frac{x}{2} \right)} = 1$$
$$\sqrt{7} \sin{\left(\frac{x}{2} \right)} = -1$$
Expand brackets in the left part

sqrt7sinx/2 = 1

This equation has no roots
Expand brackets in the left part

sqrt7sinx/2 = -1

This equation has no roots
or

do backward replacement
$$\sin{\left(\frac{x}{2} \right)} = w$$
Given the equation
$$\sin{\left(\frac{x}{2} \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$\frac{x}{2} = 2 \pi n + \operatorname{asin}{\left(w \right)}$$
$$\frac{x}{2} = 2 \pi n - \operatorname{asin}{\left(w \right)} + \pi$$
Or
$$\frac{x}{2} = 2 \pi n + \operatorname{asin}{\left(w \right)}$$
$$\frac{x}{2} = 2 \pi n - \operatorname{asin}{\left(w \right)} + \pi$$
, where n - is a integer
Divide both parts of the equation by
$$\frac{1}{2}$$
substitute w:
$$x_{1} = - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} + 2 \pi$$
$$x_{2} = 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} + 2 \pi$$
$$x_{3} = - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)}$$
$$x_{4} = 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)}$$
$$x_{1} = - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} + 2 \pi$$
$$x_{2} = 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} + 2 \pi$$
$$x_{3} = - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)}$$
$$x_{4} = 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)}$$
This roots
$$x_{3} = - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)}$$
$$x_{4} = 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)}$$
$$x_{1} = - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} + 2 \pi$$
$$x_{2} = 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} + 2 \pi$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{3}$$
For example, let's take the point
$$x_{0} = x_{3} - \frac{1}{10}$$
=
$$- 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} - \frac{1}{10}$$
=
$$- 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} - \frac{1}{10}$$
substitute to the expression
$$7 \sin^{2}{\left(\frac{x}{2} \right)} > 1$$
$$7 \sin^{2}{\left(\frac{- 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} - \frac{1}{10}}{2} \right)} > 1$$

      /         /  ___\\    
     2|1        |\/ 7 ||    
7*sin |-- + asin|-----|| > 1
      \20       \  7  //    
    

one of the solutions of our inequality is:
$$x < - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)}$$

 _____           _____           _____          
      \         /     \         /
-------ο-------ο-------ο-------ο-------
       x3      x4      x1      x2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)}$$
$$x > 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} \wedge x < - 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} + 2 \pi$$
$$x > 2 \operatorname{asin}{\left(\frac{\sqrt{7}}{7} \right)} + 2 \pi$$