We are given that $x = \frac{\pi}{6}$, and we are asked to find the exact value of $\cos(2\pi - x)$.

Step 1: Substitute the given value of $x$ into the expression.

We know that $x = \frac{\pi}{6}$, so we substitute it into $\cos(2\pi - x)$: $\cos(2\pi - \frac{\pi}{6})$

Step 2: Simplify the angle.

We simplify $2\pi - \frac{\pi}{6}$. First, express $2\pi$ with a common denominator: $2\pi = \frac{12\pi}{6}$ Now subtract $\frac{\pi}{6}$ from $\frac{12\pi}{6}$: $2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6}$

Step 3: Find the cosine of the angle $\frac{11\pi}{6}$.

The angle $\frac{11\pi}{6}$ lies in the fourth quadrant, where cosine is positive. To find the cosine, we calculate the reference angle. The reference angle is: $2\pi - \frac{11\pi}{6} = \frac{\pi}{6}$ We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$.

Since $\frac{11\pi}{6}$ is in the fourth quadrant, where cosine is positive, we have: $\cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2}$

Final Answer:

The exact value of $\cos(2\pi - x)$ is $\frac{\sqrt{3}}{2}$.

Thus, the correct answer is $\frac{\sqrt{3}}{2}$, which corresponds to: Radical 3 over 2.

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Would you like more details or have any further questions on this? Here are some related questions to explore:

1. How does the cosine function behave in different quadrants?
2. What is the reference angle for angles in the second and third quadrants?
3. Can you derive the cosine and sine of common angles like $\frac{\pi}{3}$, $\frac{\pi}{4}$, etc.?
4. How do you apply trigonometric identities to simplify expressions?
5. How do you solve for exact values of trigonometric functions at non-standard angles?

Tip: Remember that the cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants.