sin^2(x)+sin(x)<0 inequation

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

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

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

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

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

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

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

   2                      
cos (1/10) - cos(1/10) < 0
    

one of the solutions of our inequality is:
$$x < - \frac{\pi}{2}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < - \frac{\pi}{2}$$
$$x > 0 \wedge x < \pi$$
$$x > \frac{3 \pi}{2}$$