# How do you find all solutions of the equation sin^2x+sinx=0?

Jun 9, 2015

$S = \left\{x | x = k \pi \vee x = - \frac{3}{2} \pi + 2 k \pi \wedge k \in \mathbb{Z}\right\}$

#### Explanation:

${\sin}^{2} x + \sin x = 0$
can be rewritten:
$\sin x \left(\sin x + 1\right) = 0$
So we can state that $\sin x = 0$ or $\sin x = - 1$ because multiplying two numbers can result zero if and only if one of the two is zero.
So let's solve the two parts of the equation:
1. $\sin \left(x\right) = 0$
It is basically $x = k \pi$ where $k \in \mathbb{Z}$
2. $\sin \left(x\right) = - 1$
It is basically $x = - \frac{3}{2} \pi + 2 k \pi$ where $k \in \mathbb{Z}$

So the solutions are:
$S = \left\{x | x = k \pi \vee x = - \frac{3}{2} \pi + 2 k \pi \wedge k \in \mathbb{Z}\right\}$