sin^2x=<0.5 inequation

Given the inequality:
$$\sin^{2}{\left(x \right)} \leq \frac{1}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\sin^{2}{\left(x \right)} = \frac{1}{2}$$
Solve:
Given the equation
$$\sin^{2}{\left(x \right)} = \frac{1}{2}$$
transform
$$- \frac{\cos{\left(2 x \right)}}{2} = 0$$
$$\sin^{2}{\left(x \right)} - \frac{1}{2} = 0$$
Do replacement
$$w = \sin{\left(x \right)}$$
This equation is of the form

a*w^2 + b*w + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = - \frac{1}{2}$$
, then

D = b^2 - 4 * a * c = 

(0)^2 - 4 * (1) * (-1/2) = 2

Because D > 0, then the equation has two roots.

w1 = (-b + sqrt(D)) / (2*a)

w2 = (-b - sqrt(D)) / (2*a)

or
$$w_{1} = \frac{\sqrt{2}}{2}$$
$$w_{2} = - \frac{\sqrt{2}}{2}$$
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:
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(w_{1} \right)}$$
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)}$$
$$x_{1} = 2 \pi n + \frac{\pi}{4}$$
$$x_{2} = 2 \pi n + \operatorname{asin}{\left(w_{2} \right)}$$
$$x_{2} = 2 \pi n + \operatorname{asin}{\left(- \frac{\sqrt{2}}{2} \right)}$$
$$x_{2} = 2 \pi n - \frac{\pi}{4}$$
$$x_{3} = 2 \pi n - \operatorname{asin}{\left(w_{1} \right)} + \pi$$
$$x_{3} = 2 \pi n - \operatorname{asin}{\left(\frac{\sqrt{2}}{2} \right)} + \pi$$
$$x_{3} = 2 \pi n + \frac{3 \pi}{4}$$
$$x_{4} = 2 \pi n - \operatorname{asin}{\left(w_{2} \right)} + \pi$$
$$x_{4} = 2 \pi n - \operatorname{asin}{\left(- \frac{\sqrt{2}}{2} \right)} + \pi$$
$$x_{4} = 2 \pi n + \frac{5 \pi}{4}$$
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
$$x_{3} = \frac{3 \pi}{4}$$
$$x_{4} = \frac{5 \pi}{4}$$
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
$$x_{3} = \frac{3 \pi}{4}$$
$$x_{4} = \frac{5 \pi}{4}$$
This roots
$$x_{1} = - \frac{\pi}{4}$$
$$x_{2} = \frac{\pi}{4}$$
$$x_{3} = \frac{3 \pi}{4}$$
$$x_{4} = \frac{5 \pi}{4}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
=
$$- \frac{\pi}{4} - \frac{1}{10}$$
substitute to the expression
$$\sin^{2}{\left(x \right)} \leq \frac{1}{2}$$
$$\sin^{2}{\left(- \frac{\pi}{4} - \frac{1}{10} \right)} \leq \frac{1}{2}$$

   2/1    pi\       
sin |-- + --| <= 1/2
    \10   4 /       

but

   2/1    pi\       
sin |-- + --| >= 1/2
    \10   4 /       

Then
$$x \leq - \frac{\pi}{4}$$
no execute
one of the solutions of our inequality is:
$$x \geq - \frac{\pi}{4} \wedge x \leq \frac{\pi}{4}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq - \frac{\pi}{4} \wedge x \leq \frac{\pi}{4}$$
$$x \geq \frac{3 \pi}{4} \wedge x \leq \frac{5 \pi}{4}$$