https://r2.erweima.ai/mathbot/image/1725893105692jjkju4me.jpg

The equation in the image is:

\[
2 \sin^2(x) - 1 = 0
\]

### Step-by-step Solution:

1. **Rewrite the equation**:  
   \[
   2 \sin^2(x) = 1
   \]

2. **Simplify**:  
   \[
   \sin^2(x) = \frac{1}{2}
   \]

3. **Take the square root of both sides**:
   \[
   \sin(x) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}
   \]
   Therefore, \( x \) satisfies \( \sin(x) = \frac{\sqrt{2}}{2} \) or \( \sin(x) = -\frac{\sqrt{2}}{2} \).

4. **Find the general solution**:
   - For \( \sin(x) = \frac{\sqrt{2}}{2} \), the solutions are:
     \[
     x = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad x = \frac{3\pi}{4} + 2k\pi, \quad k \in \mathbb{Z}
     \]
   - For \( \sin(x) = -\frac{\sqrt{2}}{2} \), the solutions are:
     \[
     x = \frac{5\pi}{4} + 2k\pi \quad \text{or} \quad x = \frac{7\pi}{4} + 2k\pi, \quad k \in \mathbb{Z}
     \]

### Now, compare these results to the options provided:

- **Option A**: 
  \[
  \frac{\pi}{4} + 2k\pi ; \quad -\frac{\pi}{4} + 2k\pi \quad \text{(Not correct because of the \(-\frac{\pi}{4}\))}
  \]
  
- **Option B**: 
  \[
  \frac{\pi}{4} + k\pi ; \quad -\frac{\pi}{4} + k\pi \quad \text{(Not correct either)}
  \]

- **Option C**:
  \[
  \frac{\pi}{4} + k\frac{\pi}{2}; \quad -\frac{\pi}{4} + k\frac{\pi}{2}
  \]

The correct general solution is option **C**.

Would you like any further details or clarification?

### Follow-up questions:
1. What are the steps to solve trigonometric equations like this one?
2. How do we determine the general solution for sine or cosine equations?
3. What is the significance of adding \( k\pi \) or \( 2k\pi \) to the solution?
4. How can this problem be generalized for other sine and cosine values?
5. How does the unit circle help in solving trigonometric equations?

**Tip**: Always remember that trigonometric equations can have multiple solutions due to the periodicity of sine and cosine functions.

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

Find the general solution for the trigonometric equation 2sin²(x) - 1 = 0, using fundamental identities and the sine function's periodicity. A detailed explanation, suitable for Grades 10-12.

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