3sin^2x-2sinx-1≥0 inequation

Given the inequality:
$$3 \sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} - 1 \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$3 \sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} - 1 = 0$$
Solve:
Given the equation
$$3 \sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} - 1 = 0$$
transform
$$3 \sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} - 1 = 0$$
$$\left(3 \sin^{2}{\left(x \right)} - 2 \sin{\left(x \right)} - 1\right) + 0 = 0$$
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
$$a = 3$$
$$b = -2$$
$$c = -1$$
, then

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

(-2)^2 - 4 * (3) * (-1) = 16

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

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

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

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

     2                                                       
3*sin (-1/10 - asin(1/3)) - 2*sin(-1/10 - asin(1/3)) - 1 >= 0

                                    2                       
-1 + 2*sin(1/10 + asin(1/3)) + 3*sin (1/10 + asin(1/3)) >= 0
     

one of the solutions of our inequality is:
$$x \leq - \operatorname{asin}{\left(\frac{1}{3} \right)}$$

 _____           _____          
      \         /     \    
-------•-------•-------•-------
       x_3      x_1      x_2

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \operatorname{asin}{\left(\frac{1}{3} \right)}$$
$$x \geq \frac{\pi}{2} \wedge x \leq \operatorname{asin}{\left(\frac{1}{3} \right)} + \pi$$