sqrt(2sinx)<1 inequation

Given the inequality:
$$\sqrt{2 \sin{\left(x \right)}} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{2 \sin{\left(x \right)}} = 1$$
Solve:
Given the equation
$$\sqrt{2 \sin{\left(x \right)}} = 1$$
transform
$$\sqrt{2} \sqrt{\sin{\left(x \right)}} - 1 = 0$$
$$\sqrt{2 \sin{\left(x \right)}} - 1 = 0$$
Do replacement
$$w = \sin{\left(x \right)}$$
Given the equation
$$\sqrt{2} \sqrt{w} - 1 = 0$$
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:
$$\left(\sqrt{2 w}\right)^{2} = 1^{2}$$
or
$$2 w = 1$$
Divide both parts of the equation by 2

w = 1 / (2)

We get the answer: w = 1/2

The final answer:
$$w_{1} = \frac{1}{2}$$
do backward replacement
$$\sin{\left(x \right)} = w$$
Given the equation
$$\sin{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = 2 \pi n + \operatorname{asin}{\left(w \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(w \right)} + \pi$$
Or
$$x = 2 \pi n + \operatorname{asin}{\left(w \right)}$$
$$x = 2 \pi n - \operatorname{asin}{\left(w \right)} + \pi$$
, where n - is a integer
substitute w:
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(w_{1} \right)}$$
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(\frac{1}{2} \right)}$$
$$x_{1} = 2 \pi n + \frac{\pi}{6}$$
$$x_{2} = 2 \pi n - \operatorname{asin}{\left(w_{1} \right)} + \pi$$
$$x_{2} = 2 \pi n - \operatorname{asin}{\left(\frac{1}{2} \right)} + \pi$$
$$x_{2} = 2 \pi n + \frac{5 \pi}{6}$$
$$x_{1} = \frac{\pi}{6}$$
$$x_{2} = \frac{5 \pi}{6}$$
$$x_{1} = \frac{\pi}{6}$$
$$x_{2} = \frac{5 \pi}{6}$$
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
$$\sqrt{2 \sin{\left(x \right)}} < 1$$
$$\sqrt{2 \sin{\left(- \frac{1}{10} + \frac{\pi}{6} \right)}} < 1$$

          ______________    
  ___    /    /1    pi\     
\/ 2 *  /  cos|-- + --|  < 1
      \/      \10   3 /     
    

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:
$$x < \frac{\pi}{6}$$
$$x > \frac{5 \pi}{6}$$