Question

find all angles (in degrees) that satisfy the following equation.
$2\sin(\alpha) + \sqrt{3} = 0$

select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

a. the solution set is $\\{\alpha|\alpha = \square^{\circ} + k\square^{\circ}$ or $\alpha = \square^{\circ} + k\square^{\circ}\\}$, where $k$ is any integer.
(use angle measures greater than 0 and less than or equal to 360. type integers or fractions.)

b. the solution set is $\\{\alpha|\alpha = \square^{\circ} + k\square^{\circ}\\}$, where $k$ is any integer.
(use angle measures greater than 0 and less than or equal to 360. type integers or fractions.)

c. there is no solution.

Explanation:

Step1: Isolate sin(x)

$2\sin(x) = -\sqrt{3} \implies \sin(x) = -\frac{\sqrt{3}}{2}$

Step2: Find reference angle

$\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so reference angle is $60^\circ$

Step3: Identify quadrants for negative sine

Sine is negative in QIII and QIV.
QIII: $180^\circ + 60^\circ = 240^\circ$
QIV: $360^\circ - 60^\circ = 300^\circ$

Step4: Add period (360°) for all solutions

Solutions: $240^\circ + k \cdot 360^\circ$ or $300^\circ + k \cdot 360^\circ$ (k integer)

Answer:

A. The solution set is $\{x|x=240^\circ + k360^\circ \text{ or } x=300^\circ + k360^\circ\}$, where k is any integer.