The solution set of the inequality `"cos"^(2) theta lt (1)/(2)`, is

Options

A. `{theta : (8n + 1) (pi)/(4) lt theta lt (8n + 3) (pi)/(4), n in Z}`
B. `{theta: (8n-3) (pi)/(4) lt theta lt (8n-1) (pi)/(4), n in Z}`
C. `{theta: (4n + 1) (pi)/(4) lt theta lt (4n+3) (pi)/(4), n in Z}`
D. none of these

Answer

To solve the inequality \( \cos^2 \theta < \frac{1}{2} \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \cos^2 \theta < \frac{1}{2} \]

### Step 2: Take the Square Root
Taking the square root of both sides, we get:
\[ |\cos \theta| < \frac{1}{\sqrt{2}} \]
This implies:
\[ -\frac{1}{\sqrt{2}} < \cos \theta < \frac{1}{\sqrt{2}} \]

### Step 3: Find the Angles
Next, we need to find the angles where \( \cos \theta = \frac{1}{\sqrt{2}} \) and \( \cos \theta = -\frac{1}{\sqrt{2}} \).

1. For \( \cos \theta = \frac{1}{\sqrt{2}} \):
- The general solution is:
\[ \theta = 2n\pi \pm \frac{\pi}{4} \]
This gives us two specific angles:
\[ \theta = 2n\pi + \frac{\pi}{4} \quad \text{and} \quad \theta = 2n\pi - \frac{\pi}{4} \]

2. For \( \cos \theta = -\frac{1}{\sqrt{2}} \):
- The general solution is:
\[ \theta = 2n\pi \pm \frac{3\pi}{4} \]
This gives us two specific angles:
\[ \theta = 2n\pi + \frac{3\pi}{4} \quad \text{and} \quad \theta = 2n\pi - \frac{3\pi}{4} \]

### Step 4: Combine the Solutions
Now we combine the intervals from the solutions:
- From \( \cos \theta < \frac{1}{\sqrt{2}} \), we have:
\[ \theta < 2n\pi - \frac{\pi}{4} \quad \text{and} \quad \theta > 2n\pi + \frac{\pi}{4} \]
- From \( \cos \theta > -\frac{1}{\sqrt{2}} \), we have:
\[ 2n\pi - \frac{3\pi}{4} < \theta < 2n\pi + \frac{3\pi}{4} \]

### Step 5: Final Interval
The solution set can be expressed as:
\[ (2n\pi - \frac{3\pi}{4}, 2n\pi - \frac{\pi}{4}) \cup (2n\pi + \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4}) \]

### Conclusion
Thus, the solution set of the inequality \( \cos^2 \theta < \frac{1}{2} \) is:
\[ \theta \in \left(2n\pi - \frac{3\pi}{4}, 2n\pi - \frac{\pi}{4}\right) \cup \left(2n\pi + \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4}\right) \]