To solve the equation [tex]\(\cot \theta + \sqrt{3} = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex], follow these steps:
1. Rewrite the equation in terms of [tex]\(\cot \theta\)[/tex]:
[tex]\[
\cot \theta = -\sqrt{3}
\][/tex]
2. Recall the relationship between [tex]\(\cot \theta\)[/tex] and [tex]\(\tan \theta\)[/tex]:
[tex]\[
\cot \theta = \frac{1}{\tan \theta}
\][/tex]
Therefore, if [tex]\(\cot \theta = -\sqrt{3}\)[/tex], we have:
[tex]\[
\frac{1}{\tan \theta} = -\sqrt{3} \Rightarrow \tan \theta = -\frac{1}{\sqrt{3}}
\][/tex]
3. Determine the angles where [tex]\(\tan \theta = -\frac{1}{\sqrt{3}}\)[/tex]:
The value [tex]\(-\frac{1}{\sqrt{3}}\)[/tex] corresponds to the tangent values of specific angles in the unit circle.
[tex]\[
\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}
\][/tex]
Since we need a negative value, we look for angles where [tex]\(\tan \theta\)[/tex] is negative.
4. Find the specific angles within [tex]\([0, 2\pi)\)[/tex]:
[tex]\[
\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}
\][/tex]
[tex]\[
\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}
\][/tex]
5. Verify that these angles [tex]\(\theta\)[/tex] satisfy [tex]\(\tan \theta = -\frac{1}{\sqrt{3}}\)[/tex]:
- For [tex]\(\theta = \frac{5\pi}{6}\)[/tex]:
[tex]\[
\tan \left(\frac{5\pi}{6}\right) = \tan \left(\pi - \frac{\pi}{6}\right) = -\tan \left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}
\][/tex]
- For [tex]\(\theta = \frac{11\pi}{6}\)[/tex]:
[tex]\[
\tan \left(\frac{11\pi}{6}\right) = \tan \left(2\pi - \frac{\pi}{6}\right) = -\tan \left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}
\][/tex]
Therefore, the solutions to the equation [tex]\(\cot \theta + \sqrt{3} = 0\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
\theta = \frac{2\pi}{3}, \frac{5\pi}{3}
\][/tex]
