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The problem asks to find the exact value of \( s \) in the interval \( \left[\pi, \frac{3\pi}{2}\right] \) such that \( \sin(s) = -\frac{\sqrt{2}}{2} \).

### Step-by-step solution:
1. The sine function has a value of \( \sin(s) = -\frac{\sqrt{2}}{2} \) at angles where the reference angle is \( \frac{\pi}{4} \). The sine function is negative in the third and fourth quadrants.

2. The interval given is \( \left[\pi, \frac{3\pi}{2}\right] \), which corresponds to the third quadrant.

3. In the third quadrant, the angle \( s \) with a reference angle of \( \frac{\pi}{4} \) would be:
   \[
   s = \pi + \frac{\pi}{4} = \frac{5\pi}{4}
   \]

Thus, the exact value of \( s \) is \( \frac{5\pi}{4} \) radians.

Would you like more details on this, or do you have any further questions?

Here are 5 related questions to consider:
1. What is the general solution for \( \sin(s) = -\frac{\sqrt{2}}{2} \) for all quadrants?
2. Why is the sine function negative in the third and fourth quadrants?
3. How would you find the reference angle for different trigonometric functions?
4. What are the sine values of common angles like \( \frac{\pi}{6} \), \( \frac{\pi}{3} \), etc.?
5. How does the unit circle help in solving trigonometric equations?

**Tip:** Always check which quadrant the angle falls into when solving for trigonometric values within a given interval.

Find the exact value of s in the given interval [π, 3π/2] such that sin(s) = -√2/2.

Learn how to find the exact value of s in the interval [π, 3π/2] for which sin(s) equals -√2/2. This trigonometry problem covers key concepts such as the unit circle, reference angles, and sine function properties. Suitable for high school students in Grades 10-12.

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