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We are asked to solve the equation:

\[
\cos^2\theta = \frac{1}{2}
\]

on the interval \( \theta \in [0, 2\pi] \).

### Step-by-Step Solution:

1. Start by taking the square root of both sides to solve for \( \cos\theta \):
   \[
   \cos\theta = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}
   \]
   So, we have two cases to solve: \( \cos\theta = \frac{\sqrt{2}}{2} \) and \( \cos\theta = -\frac{\sqrt{2}}{2} \).

2. **For \( \cos\theta = \frac{\sqrt{2}}{2} \):**
   \[
   \theta = \frac{\pi}{4}, \quad \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}
   \]
   These are the two angles in the first and fourth quadrants where the cosine is positive.

3. **For \( \cos\theta = -\frac{\sqrt{2}}{2} \):**
   \[
   \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}, \quad \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}
   \]
   These are the two angles in the second and third quadrants where the cosine is negative.

### Final Solution:
The solutions for \( \theta \) on the interval \( [0, 2\pi] \) are:
\[
\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}
\]

Would you like any further details or clarifications?

Here are five related questions for further practice:
1. Solve \( \sin^2\theta = \frac{1}{2} \) on \( \theta \in [0, 2\pi] \).
2. Solve \( \cos\theta = \frac{1}{2} \) on \( \theta \in [0, 2\pi] \).
3. Solve \( \cos^2\theta = \frac{3}{4} \) on \( \theta \in [0, 2\pi] \).
4. Solve \( 2\cos^2\theta = 1 \) on \( \theta \in [0, 2\pi] \).
5. Solve \( \sin\theta = -\frac{1}{\sqrt{2}} \) on \( \theta \in [0, 2\pi] \).

**Tip:** When solving trigonometric equations involving squared terms, always consider both the positive and negative solutions after taking the square root.

Learn how to solve the trigonometric equation cos^2(θ) = 1/2 in the interval [0, 2π]. The solution uses the unit circle and properties of the cosine function. Step-by-step instructions are provided, suitable for Grades 9-12.

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