To solve the equation [tex]\(\cot \left(4 x - \frac{\pi}{6}\right) = \sqrt{3}\)[/tex], let's follow these steps:
1. Recognize the trigonometric identity involving [tex]\(\cot\)[/tex]:
Recall that [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex]. Given [tex]\(\cot \theta = \sqrt{3}\)[/tex], we know:
[tex]\[
\tan \theta = \frac{1}{\sqrt{3}}
\][/tex]
2. Determine the specific angles:
From trigonometric tables or the unit circle, we know:
[tex]\[
\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}
\][/tex]
Since [tex]\(\theta = \frac{\pi}{6}\)[/tex] and the tangent function is periodic with period [tex]\(\pi\)[/tex], the general solution for [tex]\(\theta\)[/tex] when [tex]\(\tan \theta = \frac{1}{\sqrt{3}}\)[/tex] is:
[tex]\[
\theta = \frac{\pi}{6} + k\pi \quad \text{for any integer } k
\][/tex]
3. Relate [tex]\(\theta\)[/tex] to our specific equation:
Here, [tex]\(\theta = 4 x - \frac{\pi}{6}\)[/tex]. Thus, we set up the equation:
[tex]\[
4 x - \frac{\pi}{6} = \frac{\pi}{6} + k\pi
\][/tex]
4. Solve for [tex]\(x\)[/tex]:
Isolate [tex]\(x\)[/tex] in the equation:
[tex]\[
4 x - \frac{\pi}{6} = \frac{\pi}{6} + k\pi
\][/tex]
Add [tex]\(\frac{\pi}{6}\)[/tex] to both sides:
[tex]\[
4 x = \frac{\pi}{6} + \frac{\pi}{6} + k\pi
\][/tex]
[tex]\[
4 x = \frac{2\pi}{6} + k\pi
\][/tex]
[tex]\[
4 x = \frac{\pi}{3} + k\pi
\][/tex]
Finally, divide by 4:
[tex]\[
x = \frac{\pi}{12} + \frac{k\pi}{4}
\][/tex]
5. Specific Solution:
For [tex]\(k = -1\)[/tex], we have:
[tex]\[
x = \frac{\pi}{12} - \frac{\pi}{4} = \frac{\pi}{12} - \frac{3\pi}{12} = -\frac{2\pi}{12} = -\frac{\pi}{6}
\][/tex]
So, the solution for the equation [tex]\(\cot \left(4 x - \frac{\pi}{6}\right) = \sqrt{3}\)[/tex] is:
[tex]\[
x = -\frac{\pi}{6}
\][/tex]
In numerical form,
[tex]\[
x \approx -0.5236
\][/tex]
Thus, the exact solution is [tex]\(x = -\frac{\pi}{6}\)[/tex], and the approximate numerical solution is [tex]\(-0.5236\)[/tex].
