We are tasked with finding the value of $s$ where $\tan(s) = -\sqrt{3}$, and $s$ is in the interval $[ \frac{3\pi}{2}, 2\pi ]$. We will use the unit circle to solve this.

Step 1: Understand the Tangent Value

The tangent function is the ratio of the sine to the cosine:

$\tan(s) = \frac{\sin(s)}{\cos(s)}$

We are given that $\tan(s) = -\sqrt{3}$. From our knowledge of the unit circle, we know that the tangent of $\frac{\pi}{3}$ is $\sqrt{3}$. So, we are looking for an angle where the tangent is $-\sqrt{3}$, which will be in the third or fourth quadrant, since the tangent is negative in those quadrants.

Step 2: Analyze the Interval

The interval $[ \frac{3\pi}{2}, 2\pi ]$ corresponds to the fourth quadrant on the unit circle. Therefore, we are looking for an angle in the fourth quadrant where the tangent is $-\sqrt{3}$.

Step 3: Identify the Angle

In the first quadrant, $\tan(\frac{\pi}{3}) = \sqrt{3}$. Since we need $\tan(s) = -\sqrt{3}$, the corresponding angle in the fourth quadrant is:

$s = 2\pi - \frac{\pi}{3} = \frac{6\pi}{3} - \frac{\pi}{3} = \frac{5\pi}{3}$

Step 4: Final Answer

Thus, the exact value of $s$ in the interval $[ \frac{3\pi}{2}, 2\pi ]$ where $\tan(s) = -\sqrt{3}$ is:

$s = \frac{5\pi}{3}$

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How do you determine the sign of tangent in different quadrants?
2. What is the relationship between sine, cosine, and tangent on the unit circle?
3. How does the unit circle help in finding exact trigonometric values?
4. Why is the tangent function positive in the first and third quadrants?
5. What is the tangent of an angle in the second quadrant?

Tip: Memorizing key angles and their trigonometric values from the unit circle can greatly speed up solving problems like these!