How did you go from step 1 and 2. Don't you have to do sin(inverse) to cancel out the sin^2(x)

Great question! Let's clarify why taking the inverse sine (\(\sin^{-1}\)) is not necessary in this particular problem.

### Steps to Solve the Equation \(\sin^2(x) - 1 = 0\)

1. **Given Equation**:  
   The original equation is:
   \[
   \sin^2(x) - 1 = 0
   \]

2. **Rewrite the Equation**:  
   To isolate \(\sin^2(x)\), add \(1\) to both sides:
   \[
   \sin^2(x) = 1
   \]

### Understanding the Transition from \(\sin^2(x) = 1\) to \(\sin(x) = \pm 1\)

- In this context, \(\sin^2(x)\) means "the square of \(\sin(x)\)". The equation \(\sin^2(x) = 1\) tells us that the square of the sine of \(x\) is equal to \(1\).
- To solve \(\sin^2(x) = 1\), we take the square root of both sides:

  \[
  \sin^2(x) = 1 \Rightarrow \sin(x) = \pm \sqrt{1}
  \]

  Simplifying further gives:

  \[
  \sin(x) = \pm 1
  \]

### Why We Don't Use \(\sin^{-1}\) (Inverse Sine)

- The inverse sine function, \(\sin^{-1}\) or \(\arcsin\), is typically used when you know the value of \(\sin(x)\) (like \(0.5\), \(-0.5\), etc.) and want to find the angle \(x\).  
- Here, we are dealing with \(\sin^2(x)\), and by taking the square root, we get direct solutions for \(\sin(x)\). Since \(\sin^2(x) = 1\) implies \(\sin(x) = \pm1\), no further step with the inverse sine function is required because the square root gives us all possible solutions.

### Finding the Specific Angles

1. **When \(\sin(x) = 1\)**:
   - This occurs at:
     \[
     x = \frac{\pi}{2}
     \]

2. **When \(\sin(x) = -1\)**:
   - This occurs at:
     \[
     x = \frac{3\pi}{2}
     \]

### Final Solution

The solutions to the equation \(\sin^2(x) - 1 = 0\) in the interval \(0 \leq x \leq 2\pi\) are:

\[
x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2}
\]

Would you like more details or have any further questions?

### Related Questions:
1. How would you solve \( \cos^2(x) = \frac{1}{2} \) for \( 0 \leq x \leq 2\pi \)?
2. What is the general solution for \( \sin^2(x) = \frac{1}{4} \)?
3. How do you solve the equation \( 2\sin^2(x) - 1 = 0 \)?
4. What values of \( x \) satisfy \( \sin(3x) = 0 \) within the given interval?
5. Can you derive solutions to \( \tan^2(x) = 1 \) for \( 0 \leq x \leq 2\pi \)?

### Tip:
When solving trigonometric equations, use identities such as \( \sin^2(x) + \cos^2(x) = 1 \) and think about the specific values of angles that satisfy the conditions in the unit circle.

Explore the solution to the trigonometric equation \(\sin^2(x) - 1 = 0\) step-by-step. Learn why taking the inverse sine (\(\sin^{-1}\)) is not required, and find specific solutions within the interval \(0 \leq x \leq 2\pi\). Suitable for advanced high school students and anyone interested in trigonometric identities and equations.

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