# Find all value of α in [0°, 360°) that will satisfy this equation? 2sinα=-√3

Apr 8, 2018

$\alpha = {240}^{o} , {300}^{o}$

#### Explanation:

Isolate the sine.

$2 \sin \alpha = - \sqrt{3}$

$\sin \alpha = - \frac{\sqrt{3}}{2}$

Consider where, in the interval $\left[{0}^{o} , {360}^{o}\right) ,$ the sine function equals $- \frac{\sqrt{3}}{2}$. Our solutions will have to be in the third and fourth quadrants, as these are the quadrants in which the sine is negative.

So, from the unit circle itself, we get

$\alpha = {240}^{o} , {300}^{o}$

Apr 8, 2018

alpha= 240°, 300°

#### Explanation:

$2 \sin \alpha = - \sqrt{3}$

Divide by 2:
$\sin \alpha = - \frac{\sqrt{3}}{2}$

Look at the unit circle to see when y is equal to $- \frac{\sqrt{3}}{2}$:

alpha= 240°, 300°

graph{2sinx+sqrt3 [-10, 10, -5, 5]}