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To solve the equation \( 2 \cos^2 \theta^\circ = 1 - \sin \theta^\circ \) for \( 0 \leq \theta \leq 360^\circ \), we can follow these steps:

### Step-by-Step Solution

1. **Rewrite \( \cos^2 \theta \) in terms of \( \sin \theta \):**

   Using the Pythagorean identity \( \cos^2 \theta = 1 - \sin^2 \theta \), substitute this into the equation:
   \[
   2(1 - \sin^2 \theta) = 1 - \sin \theta.
   \]
   
2. **Expand and simplify:**
   \[
   2 - 2 \sin^2 \theta = 1 - \sin \theta.
   \]
   Move all terms to one side to set the equation to zero:
   \[
   2 \sin^2 \theta - \sin \theta - 1 = 0.
   \]

3. **Solve the quadratic equation in \( \sin \theta \):**

   This is a quadratic equation in terms of \( \sin \theta \), where \( a = 2 \), \( b = -1 \), and \( c = -1 \):
   \[
   2 \sin^2 \theta - \sin \theta - 1 = 0.
   \]
   Use the quadratic formula:
   \[
   \sin \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}.
   \]
   Simplify inside the square root:
   \[
   \sin \theta = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}.
   \]
   This gives two solutions:
   \[
   \sin \theta = \frac{1 + 3}{4} = 1 \quad \text{and} \quad \sin \theta = \frac{1 - 3}{4} = -\frac{1}{2}.
   \]

4. **Find \( \theta \) for each case:**

   - **Case 1:** \( \sin \theta = 1 \)
     - \( \theta = 90^\circ \).
   
   - **Case 2:** \( \sin \theta = -\frac{1}{2} \)
     - For \( \sin \theta = -\frac{1}{2} \), the angles are \( \theta = 210^\circ \) and \( \theta = 330^\circ \) in the range \( 0 \leq \theta \leq 360^\circ \).

### Final Solutions
The values of \( \theta \) that satisfy the equation are:
\[
\theta = 90^\circ, \quad \theta = 210^\circ, \quad \theta = 330^\circ.
\]

Would you like further details on any step?

### Related Questions
1. How do we apply trigonometric identities to simplify equations?
2. What other trigonometric equations can be solved by converting to a quadratic form?
3. How can we verify if a solution to a trigonometric equation is correct?
4. How do we handle trigonometric equations with restricted domains?
5. Can this approach be used for equations with functions like \( \tan \theta \) or \( \sec \theta \)?

### Tip
When solving trigonometric equations, always remember to check if the solutions fall within the given range or interval for the variable.

Solve the equation 2 cos² θ° = 1 - sin θ° for 0 ≤ θ ≤ 360°.

Step-by-step solution to solving the trigonometric equation 2 cos² θ° = 1 - sin θ° for 0 ≤ θ ≤ 360°. This problem covers trigonometric identities and quadratic equations, suitable for students in Grades 10-12.

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