sgrt2sin(pi/2-2x)>1 inequation

Given the inequality:
$$\sqrt{2 \sin{\left(- 2 x + \frac{\pi}{2} \right)}} > 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{2 \sin{\left(- 2 x + \frac{\pi}{2} \right)}} = 1$$
Solve:
Given the equation
$$\sqrt{2 \sin{\left(- 2 x + \frac{\pi}{2} \right)}} = 1$$
transform
$$\sqrt{2} \sqrt{\cos{\left(2 x \right)}} - 1 = 0$$
$$\sqrt{2 \sin{\left(- 2 x + \frac{\pi}{2} \right)}} - 1 = 0$$
Do replacement
$$w = \cos{\left(2 x \right)}$$
Given the equation
$$\sqrt{2 \sin{\left(- 2 x + \frac{\pi}{2} \right)}} - 1 = 0$$
Because equation degree is equal to = 1/2 - does not contain even numbers in the numerator, then
the equation has single real root.
We raise the equation sides to 2-th degree:
We get:
$$\left(\sqrt{2}\right)^{2} \left(\sqrt{0 w + \cos{\left(2 x \right)}}\right)^{2} = 1^{2}$$
or
$$2 \cos{\left(2 x \right)} = 1$$
Expand brackets in the left part

2*cos2*x = 1

This equation has no roots

do backward replacement
$$\cos{\left(2 x \right)} = w$$
Given the equation
$$\cos{\left(2 x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$2 x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$2 x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$2 x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$2 x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
Divide both parts of the equation by
$$2$$
substitute w:
$$x_{1} = \frac{\pi}{6}$$
$$x_{2} = \frac{5 \pi}{6}$$
$$x_{1} = \frac{\pi}{6}$$
$$x_{2} = \frac{5 \pi}{6}$$
This roots
$$x_{1} = \frac{\pi}{6}$$
$$x_{2} = \frac{5 \pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{6}$$
=
$$- \frac{1}{10} + \frac{\pi}{6}$$
substitute to the expression
$$\sqrt{2 \sin{\left(- 2 x + \frac{\pi}{2} \right)}} > 1$$
$$\sqrt{2 \sin{\left(- 2 \left(- \frac{1}{10} + \frac{\pi}{6}\right) + \frac{\pi}{2} \right)}} > 1$$

          _____________    
  ___    /    /1   pi\     
\/ 2 *  /  sin|- + --|  > 1
      \/      \5   6 /     
    

one of the solutions of our inequality is:
$$x < \frac{\pi}{6}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < \frac{\pi}{6}$$
$$x > \frac{5 \pi}{6}$$