# How do you solve sin^2x+sinx=0 and find all solutions in the interval [0,2pi)?

Sep 23, 2016

$x = 0 , \frac{3 \pi}{2} , \pi$

#### Explanation:

We have: ${\sin}^{2} \left(x\right) + \sin \left(x\right) = 0$; $\left[0 , 2 \pi\right)$

$\implies \sin \left(x\right) \left(\sin \left(x\right) + 1\right) = 0$

$\implies \sin \left(x\right) = 0$

$\implies x = 0 , \left(\pi - 0\right) , \left(\pi + 0\right) , \left(2 \pi - 0\right)$

or

$\implies \sin \left(x\right) + 1 = 0$

$\implies \sin \left(x\right) = - 1$

$\implies x = \pi + \frac{\pi}{2}$

$\implies x = 0 , \frac{3 \pi}{2} , \pi$