To solve the equation [tex]\(\sqrt{3} \csc \theta + 2 = 0\)[/tex] for [tex]\(\theta\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex], follow this detailed, step-by-step solution:
1. Isolate the cosecant function:
Given:
[tex]\[
\sqrt{3} \csc \theta + 2 = 0
\][/tex]
Subtract 2 from both sides:
[tex]\[
\sqrt{3} \csc \theta = -2
\][/tex]
Divide both sides by [tex]\(\sqrt{3}\)[/tex]:
[tex]\[
\csc \theta = -\frac{2}{\sqrt{3}}
\][/tex]
Recognize that the cosecant function is the reciprocal of the sine function:
[tex]\[
\csc \theta = \frac{1}{\sin \theta}
\][/tex]
Therefore:
[tex]\[
\frac{1}{\sin \theta} = -\frac{2}{\sqrt{3}}
\][/tex]
2. Solve for [tex]\(\sin \theta\)[/tex]:
Take the reciprocal of both sides:
[tex]\[
\sin \theta = -\frac{\sqrt{3}}{2}
\][/tex]
3. Determine where [tex]\(\sin \theta = -\frac{\sqrt{3}}{2}\)[/tex]:
The sine function [tex]\(\sin \theta\)[/tex] achieves the value [tex]\(-\frac{\sqrt{3}}{2}\)[/tex] at specific angles within the interval [tex]\([0, 2\pi)\)[/tex]:
- Recognize that [tex]\(\sin \theta = -\frac{\sqrt{3}}{2}\)[/tex] corresponds to the angles where the sine value is negative and has the corresponding reference angle of [tex]\(\frac{\pi}{3}\)[/tex].
- The angles in the interval [tex]\([0, 2\pi)\)[/tex] where this value occurs are in the third and fourth quadrants:
- Third quadrant: [tex]\(\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\)[/tex]
- Fourth quadrant: [tex]\(\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\)[/tex]
4. Write the solutions:
The solutions to the equation [tex]\(\sqrt{3} \csc \theta + 2 = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
\theta = \frac{4\pi}{3}, \frac{5\pi}{3}
\][/tex]
So, the solutions in terms of [tex]\(\pi\)[/tex] are:
[tex]\[
\theta = \frac{4\pi}{3}, \frac{5\pi}{3}
\][/tex]
