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Let's solve the equation step by step:

The given equation is:
\[
2 \sin^2(x) - 1 = 0
\]

### Step 1: Solve for \(\sin^2(x)\)
\[
2 \sin^2(x) = 1
\]
\[
\sin^2(x) = \frac{1}{2}
\]

### Step 2: Solve for \(\sin(x)\)
\[
\sin(x) = \pm \frac{\sqrt{2}}{2}
\]

### Step 3: Find the general solutions for \(\sin(x) = \frac{\sqrt{2}}{2}\) and \(\sin(x) = -\frac{\sqrt{2}}{2}\)

For \(\sin(x) = \frac{\sqrt{2}}{2}\), the general solutions are:
\[
x = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad x = \frac{3\pi}{4} + 2k\pi \quad \text{where } k \in \mathbb{Z}
\]

For \(\sin(x) = -\frac{\sqrt{2}}{2}\), the general solutions are:
\[
x = \frac{5\pi}{4} + 2k\pi \quad \text{or} \quad x = \frac{7\pi}{4} + 2k\pi \quad \text{where } k \in \mathbb{Z}
\]

### Step 4: Match the solution with the given options

- **Option A**: \(\frac{\pi}{4} + 2k\pi\) and \(-\frac{\pi}{4} + 2k\pi\) — this doesn't match our solutions.
- **Option B**: \(\frac{\pi}{4} + k\pi\) and \(-\frac{\pi}{4} + k\pi\) — this corresponds to all possible solutions for \(\sin(x) = \pm\frac{\sqrt{2}}{2}\), since \(k\pi\) covers all the cases.
- **Option C**: \(\frac{\pi}{4} + \frac{k\pi}{2}\) and \(-\frac{\pi}{4} + \frac{k\pi}{2}\) — this doesn't match the correct form.

### Conclusion:
The correct answer is **B**.

Would you like more details or have any questions about the solution? Here are five related questions to deepen your understanding:

1. How do you find the general solution of trigonometric equations involving sine or cosine?
2. What are the typical steps for solving quadratic trigonometric equations?
3. How does the period of the sine function affect the general solution?
4. Can you derive the solutions for \(\sin(x) = \frac{\sqrt{2}}{2}\) using a unit circle?
5. What is the importance of adding \(2k\pi\) (or \(k\pi\)) when solving trigonometric equations?

**Tip:** Always check the periodic nature of trigonometric functions to ensure you find all possible solutions in the general form.

Quelle est la solution dans ℝ de l'équation suivante : 2sin²(x) - 1 = 0 ?

Learn how to solve the trigonometric equation 2sin²(x) - 1 = 0. This problem involves using the Pythagorean identity and finding the general solution for sine equations. Suitable for high school students in Grades 10-12.

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