4sin^2x+8sinx+3≤0 inequation

Given the inequality:
$$4 \sin^{2}{\left(x \right)} + 8 \sin{\left(x \right)} + 3 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$4 \sin^{2}{\left(x \right)} + 8 \sin{\left(x \right)} + 3 = 0$$
Solve:
Given the equation
$$4 \sin^{2}{\left(x \right)} + 8 \sin{\left(x \right)} + 3 = 0$$
transform
$$4 \sin^{2}{\left(x \right)} + 8 \sin{\left(x \right)} + 3 = 0$$
$$\left(4 \sin^{2}{\left(x \right)} + 8 \sin{\left(x \right)} + 3\right) + 0 = 0$$
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
$$a = 4$$
$$b = 8$$
$$c = 3$$
, then

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

(8)^2 - 4 * (4) * (3) = 16

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

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

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

or
$$w_{1} = - \frac{1}{2}$$
Simplify
$$w_{2} = - \frac{3}{2}$$
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:
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(w_{1} \right)}$$
$$x_{1} = 2 \pi n + \operatorname{asin}{\left(- \frac{1}{2} \right)}$$
$$x_{1} = 2 \pi n - \frac{\pi}{6}$$
$$x_{2} = 2 \pi n + \operatorname{asin}{\left(w_{2} \right)}$$
$$x_{2} = 2 \pi n + \operatorname{asin}{\left(- \frac{3}{2} \right)}$$
$$x_{2} = 2 \pi n - \operatorname{asin}{\left(\frac{3}{2} \right)}$$
$$x_{3} = 2 \pi n - \operatorname{asin}{\left(w_{1} \right)} + \pi$$
$$x_{3} = 2 \pi n - \operatorname{asin}{\left(- \frac{1}{2} \right)} + \pi$$
$$x_{3} = 2 \pi n + \frac{7 \pi}{6}$$
$$x_{4} = 2 \pi n - \operatorname{asin}{\left(w_{2} \right)} + \pi$$
$$x_{4} = 2 \pi n + \pi - \operatorname{asin}{\left(- \frac{3}{2} \right)}$$
$$x_{4} = 2 \pi n + \pi + \operatorname{asin}{\left(\frac{3}{2} \right)}$$
$$x_{1} = - \frac{\pi}{6}$$
$$x_{2} = \frac{7 \pi}{6}$$
$$x_{3} = \pi + \operatorname{asin}{\left(\frac{3}{2} \right)}$$
$$x_{4} = - \operatorname{asin}{\left(\frac{3}{2} \right)}$$
Exclude the complex solutions:
$$x_{1} = - \frac{\pi}{6}$$
$$x_{2} = \frac{7 \pi}{6}$$
This roots
$$x_{1} = - \frac{\pi}{6}$$
$$x_{2} = \frac{7 \pi}{6}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{\pi}{6} - \frac{1}{10}$$
=
$$- \frac{\pi}{6} - \frac{1}{10}$$
substitute to the expression
$$4 \sin^{2}{\left(x \right)} + 8 \sin{\left(x \right)} + 3 \leq 0$$
$$8 \sin{\left(- \frac{\pi}{6} - \frac{1}{10} \right)} + 4 \sin^{2}{\left(- \frac{\pi}{6} - \frac{1}{10} \right)} + 3 \leq 0$$

         /1    pi\        2/1    pi\     
3 - 8*sin|-- + --| + 4*sin |-- + --| <= 0
         \10   6 /         \10   6 /     

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

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \frac{\pi}{6}$$
$$x \geq \frac{7 \pi}{6}$$