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The given equation is:

\[
\sin^2 \theta = 1
\]

We need to solve this equation for \( \theta \) in the interval \( 0^\circ \leq \theta < 360^\circ \).

### Step 1: Solve for \( \sin \theta \)
First, take the square root of both sides of the equation:

\[
\sin \theta = \pm 1
\]

So, \( \sin \theta = 1 \) or \( \sin \theta = -1 \).

### Step 2: Solve for \( \theta \) where \( \sin \theta = 1 \)

The sine function equals 1 at:
\[
\theta = 90^\circ
\]

### Step 3: Solve for \( \theta \) where \( \sin \theta = -1 \)

The sine function equals -1 at:
\[
\theta = 270^\circ
\]

### Final Solution:

Therefore, the solutions to the equation \( \sin^2 \theta = 1 \) within the interval \( 0^\circ \leq \theta < 360^\circ \) are:
\[
\theta = 90^\circ, 270^\circ
\]

Would you like further details on any step?

Here are five related questions to further your understanding:
1. What is the general solution for \( \sin^2 \theta = 1 \) without the restricted interval?
2. How can the unit circle be used to visualize solutions for \( \sin^2 \theta = 1 \)?
3. What are the periodic properties of the sine function?
4. How does the Pythagorean identity relate to \( \sin^2 \theta = 1 \)?
5. Can we solve similar equations for other trigonometric functions, like \( \cos^2 \theta = 1 \)?

**Tip:** Remember that trigonometric equations often have multiple solutions within a given interval because of the periodic nature of the sine and cosine functions.

Solve sin^2(θ) = 1 given that 0° ≤ θ < 360°.

Learn how to solve the trigonometric equation sin^2(θ) = 1 for 0° ≤ θ < 360°. This solution covers key trigonometric concepts, such as the sine function's behavior on the unit circle. Suitable for students in Grades 9-12.

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