# How do I solve the following equation? The answers that I came up with are wrong

## solve the given equation: $\sin \theta \left(2 \sin \theta + 1\right) = 0$

Feb 9, 2018

In the interval $\left[0 , 2 \pi\right)$, $\theta = \left\{0 , \pi , \frac{5 \pi}{6} , \frac{7 \pi}{6}\right\}$

#### Explanation:

There are two factors in your equation, basically $\sin \theta \times \left(2 \sin \theta + 1\right)$. If their product is zero, then one or both factors must be equal to zero as well (zero product property). Set each factor equal to zero and solve:

$\sin \theta = 0$

$\sin \theta$ is zero at 0 and $\pi$ (look at your unit circle)

$2 \sin \theta + 1 = 0$
$2 \sin \theta = - 1$
$\sin \theta = - \frac{1}{2}$

$\sin \theta$ $- \frac{1}{2}$ at $\frac{5 \pi}{6}$ and $\frac{7 \pi}{6}$ (again, look at your unit circle)

If you are looking for a general solution, you would add $\pi n$ to 0 and $2 \pi n$ to $\frac{5 \pi}{6}$ and $\frac{7 \pi}{6}$

Hope this helps.