Answer:
[tex]\displaystyle x=\left\{n\pi, \frac{3\pi}{2}+2n\pi \right\}, n\in\mathbb{Z}[/tex]
Step-by-step explanation:
We want to solve the equation:
[tex]\sin^2x+\sin x=0[/tex]
Factor:
[tex]\sin x(\sin x+1)=0[/tex]
Zero Product Property:
[tex]\sin x=0\text{ or } \sin x+1=0[/tex]
Recall that sine equals 0 at 0 radians. This will repeat every π radians. So:
[tex]x=n\pi, n\in\mathbb{Z}[/tex]
(Where n is an integer).
In the second case, we have:
[tex]\sin x=-1[/tex]
This occurs at 3π/2 and will repeat every cycle or 2π. Hence:
[tex]\displaystyle x=\frac{3\pi}{2}+2n\pi, n\in\mathbb{Z}[/tex]
Our solution is:
[tex]\displaystyle x=\left\{n\pi, \frac{3\pi}{2}+2n\pi \right\}, n\in\mathbb{Z}[/tex]
Notes:
In the interval [0, 2π), the solutions are:
[tex]\displaystyle x=\left\{0,\pi, \frac{3\pi}{2}\right\}[/tex]
