cos^2(x)-1/2cosx<=0 inequation

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

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

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

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

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

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

or
$$w_{1} = \frac{1}{2}$$
$$w_{2} = 0$$
do backward replacement
$$\cos{\left(x \right)} = w$$
Given the equation
$$\cos{\left(x \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$x = \pi n + \operatorname{acos}{\left(w \right)}$$
$$x = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
substitute w:
$$x_{1} = \pi n + \operatorname{acos}{\left(w_{1} \right)}$$
$$x_{1} = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$x_{1} = \pi n + \frac{\pi}{3}$$
$$x_{2} = \pi n + \operatorname{acos}{\left(w_{2} \right)}$$
$$x_{2} = \pi n + \operatorname{acos}{\left(0 \right)}$$
$$x_{2} = \pi n + \frac{\pi}{2}$$
$$x_{3} = \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi$$
$$x_{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$x_{3} = \pi n - \frac{2 \pi}{3}$$
$$x_{4} = \pi n + \operatorname{acos}{\left(w_{2} \right)} - \pi$$
$$x_{4} = \pi n - \pi + \operatorname{acos}{\left(0 \right)}$$
$$x_{4} = \pi n - \frac{\pi}{2}$$
$$x_{1} = \frac{\pi}{3}$$
$$x_{2} = \frac{\pi}{2}$$
$$x_{3} = \frac{3 \pi}{2}$$
$$x_{4} = \frac{5 \pi}{3}$$
$$x_{1} = \frac{\pi}{3}$$
$$x_{2} = \frac{\pi}{2}$$
$$x_{3} = \frac{3 \pi}{2}$$
$$x_{4} = \frac{5 \pi}{3}$$
This roots
$$x_{1} = \frac{\pi}{3}$$
$$x_{2} = \frac{\pi}{2}$$
$$x_{3} = \frac{3 \pi}{2}$$
$$x_{4} = \frac{5 \pi}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{3}$$
=
$$- \frac{1}{10} + \frac{\pi}{3}$$
substitute to the expression
$$\cos^{2}{\left(x \right)} - \frac{\cos{\left(x \right)}}{2} \leq 0$$
$$- \frac{\cos{\left(- \frac{1}{10} + \frac{\pi}{3} \right)}}{2} + \cos^{2}{\left(- \frac{1}{10} + \frac{\pi}{3} \right)} \leq 0$$

                   /1    pi\     
                sin|-- + --|     
   2/1    pi\      \10   6 / <= 0
sin |-- + --| - ------------     
    \10   6 /        2           

but

                   /1    pi\     
                sin|-- + --|     
   2/1    pi\      \10   6 / >= 0
sin |-- + --| - ------------     
    \10   6 /        2           

Then
$$x \leq \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$x \geq \frac{\pi}{3} \wedge x \leq \frac{\pi}{2}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq \frac{\pi}{3} \wedge x \leq \frac{\pi}{2}$$
$$x \geq \frac{3 \pi}{2} \wedge x \leq \frac{5 \pi}{3}$$