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The given problem is:

Solve \(\sin^2 x - 2 \sin x - 3 = 0\), where \( x \in [0, 2\pi] \). (Hint: Let \( t = \sin x \) for change of variable)

Let's solve this step-by-step:

1. **Substitute** \( t = \sin x \) into the equation. The equation becomes:
   \[
   t^2 - 2t - 3 = 0
   \]

2. **Solve the quadratic equation** \( t^2 - 2t - 3 = 0 \):
   \[
   t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 1 \), \( b = -2 \), and \( c = -3 \). Plugging in these values:
   \[
   t = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)}
   \]
   \[
   t = \frac{2 \pm \sqrt{4 + 12}}{2}
   \]
   \[
   t = \frac{2 \pm \sqrt{16}}{2}
   \]
   \[
   t = \frac{2 \pm 4}{2}
   \]
   So, the solutions for \( t \) are:
   \[
   t = 3 \quad \text{and} \quad t = -1
   \]

3. **Check for valid values of \( t \)**:
   \(\sin x\) can only take values between \(-1\) and \(1\). So, \( t = 3 \) is not valid. The only valid value is:
   \[
   t = -1
   \]

4. **Find \( x \) for \( \sin x = -1 \)**:
   \[
   x = \frac{3\pi}{2}
   \]
   in the interval \([0, 2\pi]\).

So, the solution is:
\[
x = \frac{3\pi}{2}
\]

Would you like more details or have any other questions?

### Related Questions:
1. How do you solve quadratic equations using the quadratic formula?
2. What are the general values of \( x \) for which \( \sin x = k \)?
3. How do you determine if a value is valid for trigonometric functions?
4. Can the same approach be used to solve other trigonometric equations?
5. What is the significance of the interval \( [0, 2\pi] \) in trigonometry?

### Tip:
Always ensure that the solutions for trigonometric equations fall within the specified range to confirm their validity.

Solve sin^2(x) - 2sin(x) - 3 = 0, where x ∈ [0, 2π]. (Hint: Let t = sin(x) for change of variable)

Learn how to solve the trigonometric quadratic equation sin^2(x) - 2sin(x) - 3 = 0 within the interval [0, 2π]. This step-by-step guide covers key concepts in trigonometry and quadratic equations, using the quadratic formula for solution.

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