# How do you solve 2sin^2x + sinx =0 from [0,2pi]?

Jul 25, 2015

Solve $2 {\sin}^{2} x + \sin x = 0$ $\left[0 , 2 \pi\right]$

#### Explanation:

$\sin x \left(2 \sin x + 1\right) = 0$
Use trig table and unit circle -->

a. sin x = 0 => x = 0, and $x = \pi$ and $x = 2 \pi$

b. 2sin x + 1 = 0 => $\sin x = - \frac{1}{2}$ => $x = \frac{7 \pi}{6}$ and $x = \frac{11 \pi}{6}$.

Finally, within interval $\left[0 , 2 \pi\right] ,$ there are 5 answers:
$0 , \pi , 2 \pi , \frac{7 \pi}{6} \mathmr{and} \frac{11 \pi}{6}$