To find all solutions of the equation [tex]\(\sqrt{3} \csc \theta + 2 = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex], follow these steps:
1. Rewrite the equation:
[tex]\[\sqrt{3} \csc \theta + 2 = 0\][/tex]
The cosecant function, [tex]\(\csc \theta\)[/tex], is the reciprocal of the sine function, [tex]\(\sin \theta\)[/tex]:
[tex]\[\csc \theta = \frac{1}{\sin \theta}\][/tex]
2. Substitute [tex]\(\csc \theta\)[/tex] with [tex]\(\frac{1}{\sin \theta}\)[/tex]:
[tex]\[\sqrt{3} \left(\frac{1}{\sin \theta}\right) + 2 = 0\][/tex]
3. Simplify the equation:
[tex]\[\frac{\sqrt{3}}{\sin \theta} + 2 = 0\][/tex]
4. Isolate the sine function:
[tex]\[\frac{\sqrt{3}}{\sin \theta} = -2\][/tex]
5. Express [tex]\(\sin \theta\)[/tex] in terms of known values:
[tex]\[\sin \theta = -\frac{\sqrt{3}}{2}\][/tex]
6. Find [tex]\(\theta\)[/tex] for [tex]\(\sin \theta = -\frac{\sqrt{3}}{2}\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex]:
The reference angle for [tex]\(\sin^{-1}(-\frac{\sqrt{3}}{2})\)[/tex] is [tex]\(\frac{\pi}{3}\)[/tex]. The sine function is negative in the third and fourth quadrants. Therefore, the solutions will be:
[tex]\[
\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}
\][/tex]
[tex]\[
\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}
\][/tex]
7. Write the final solutions:
[tex]\[
\theta = \frac{4\pi}{3}, \frac{5\pi}{3}
\][/tex]
Thus, the solutions in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
\theta = \frac{4\pi}{3}, \frac{5\pi}{3}
\][/tex]
