Find x if `2sin^(2)x -1=0`

Options

A. `""_(4)^(pi)`
B. `""_(2)^(pi)`
C. 0
D. `pi`

Answer

To solve the equation \( 2\sin^2 x - 1 = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
Start with the given equation:
\[ 2\sin^2 x - 1 = 0 \]
Add 1 to both sides:
\[ 2\sin^2 x = 1 \]

### Step 2: Divide by 2
Now, divide both sides by 2:
\[ \sin^2 x = \frac{1}{2} \]

### Step 3: Take the square root
Next, take the square root of both sides. Remember to consider both the positive and negative roots:
\[ \sin x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \]

### Step 4: Find the angles
Now, we need to find the angles \( x \) for which \( \sin x = \frac{\sqrt{2}}{2} \) and \( \sin x = -\frac{\sqrt{2}}{2} \).

1. For \( \sin x = \frac{\sqrt{2}}{2} \):
- The angles are:
\[ x = 45^\circ + 360^\circ n \quad \text{or} \quad x = 135^\circ + 360^\circ n \quad (n \in \mathbb{Z}) \]
- In radians, this is:
\[ x = \frac{\pi}{4} + 2\pi n \quad \text{or} \quad x = \frac{3\pi}{4} + 2\pi n \]

2. For \( \sin x = -\frac{\sqrt{2}}{2} \):
- The angles are:
\[ x = 225^\circ + 360^\circ n \quad \text{or} \quad x = 315^\circ + 360^\circ n \quad (n \in \mathbb{Z}) \]
- In radians, this is:
\[ x = \frac{5\pi}{4} + 2\pi n \quad \text{or} \quad x = \frac{7\pi}{4} + 2\pi n \]

### Final Answer
Thus, the general solutions for \( x \) are:
\[ x = \frac{\pi}{4} + 2\pi n, \quad x = \frac{3\pi}{4} + 2\pi n, \quad x = \frac{5\pi}{4} + 2\pi n, \quad x = \frac{7\pi}{4} + 2\pi n \quad (n \in \mathbb{Z}) \]