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The problem asks to find \( \cot 2x \) given that \( \cos x = \frac{5}{6} \) with \( x \) in the fourth quadrant (QIV).

Let's solve this step-by-step.

### Step 1: Identify \( \sin x \)
Since \( x \) is in QIV, \( \sin x \) will be negative. We can find \( \sin x \) using the Pythagorean identity:
\[
\sin^2 x + \cos^2 x = 1
\]
Substitute \( \cos x = \frac{5}{6} \):
\[
\sin^2 x + \left(\frac{5}{6}\right)^2 = 1
\]
\[
\sin^2 x + \frac{25}{36} = 1
\]
\[
\sin^2 x = 1 - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36}
\]
\[
\sin x = -\sqrt{\frac{11}{36}} = -\frac{\sqrt{11}}{6}
\]
(since \( \sin x \) is negative in QIV).

### Step 2: Find \( \cot 2x \)
We use the double-angle formula for cotangent:
\[
\cot 2x = \frac{\cos^2 x - \sin^2 x}{2 \cos x \sin x}
\]
Substitute \( \cos x = \frac{5}{6} \) and \( \sin x = -\frac{\sqrt{11}}{6} \):
\[
\cos^2 x = \left(\frac{5}{6}\right)^2 = \frac{25}{36}
\]
\[
\sin^2 x = \left(-\frac{\sqrt{11}}{6}\right)^2 = \frac{11}{36}
\]
\[
\cot 2x = \frac{\frac{25}{36} - \frac{11}{36}}{2 \cdot \frac{5}{6} \cdot -\frac{\sqrt{11}}{6}}
\]
\[
= \frac{\frac{14}{36}}{\frac{-10\sqrt{11}}{36}}
\]
\[
= \frac{14}{-10\sqrt{11}}
\]
Simplify by rationalizing:
\[
\cot 2x = \frac{14}{-10\sqrt{11}} \cdot \frac{\sqrt{11}}{\sqrt{11}} = -\frac{14\sqrt{11}}{110} = -\frac{7\sqrt{11}}{55}
\]

### Final Answer
\[
\cot 2x = -\frac{7\sqrt{11}}{55}
\]

Would you like further explanation on any step, or need additional details?

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Here are 5 follow-up questions related to this problem:
1. How would you find \( \tan x \) given \( \cos x \) and the quadrant information?
2. What would \( \sin 2x \) and \( \cos 2x \) be in this context?
3. How would the answer change if \( x \) were in the second quadrant instead?
4. What is the significance of rationalizing the denominator in trigonometric simplifications?
5. How does the double-angle formula for sine and cosine help in solving similar problems?

**Tip:** When dealing with trigonometric identities, always remember to check the quadrant to ensure the correct sign for sine and cosine values.

Let cos x = 5/6 with x in QIV and find cot 2x.

Learn how to solve for cot 2x given that cos x = 5/6 with x in the fourth quadrant. This problem covers essential trigonometric concepts, including double-angle identities and quadrant analysis, ideal for students in Grades 10-12.

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