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

Options

A. `{theta : (8n + 1) (pi)/(4) lt theta (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
Starting with the given inequality:
\[ \cos^2 \theta < \frac{1}{2} \]
We can take the square root of both sides, keeping in mind that the cosine function can be both positive and negative:
\[ -\frac{1}{\sqrt{2}} < \cos \theta < \frac{1}{\sqrt{2}} \]

### Step 2: Determine the angles corresponding to the cosine values
The values of \( \cos \theta \) correspond to specific angles in the unit circle. We need to find the angles where:
\[ \cos \theta = \frac{1}{\sqrt{2}} \quad \text{and} \quad \cos \theta = -\frac{1}{\sqrt{2}} \]
The angles for \( \cos \theta = \frac{1}{\sqrt{2}} \) are:
\[ \theta = \frac{\pi}{4} + 2n\pi \quad \text{and} \quad \theta = -\frac{\pi}{4} + 2n\pi \]
The angles for \( \cos \theta = -\frac{1}{\sqrt{2}} \) are:
\[ \theta = \frac{3\pi}{4} + 2n\pi \quad \text{and} \quad \theta = -\frac{3\pi}{4} + 2n\pi \]

### Step 3: Set up the intervals
From the values we found, we can set up the intervals:
1. For \( \cos \theta < \frac{1}{\sqrt{2}} \):
- The interval is:
\[ \frac{\pi}{4} + 2n\pi < \theta < \frac{3\pi}{4} + 2n\pi \]
2. For \( \cos \theta > -\frac{1}{\sqrt{2}} \):
- The interval is:
\[ -\frac{3\pi}{4} + 2n\pi < \theta < -\frac{\pi}{4} + 2n\pi \]

### Step 4: Combine the intervals
Combining these intervals gives us the solution set for the inequality:
1. From the first case:
\[ 2n\pi + \frac{\pi}{4} < \theta < 2n\pi + \frac{3\pi}{4} \]
2. From the second case:
\[ 2n\pi - \frac{3\pi}{4} < \theta < 2n\pi - \frac{\pi}{4} \]

### Final Result
Thus, the complete solution set for the inequality \( \cos^2 \theta < \frac{1}{2} \) is:
\[ \left( 2n\pi + \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4} \right) \cup \left( 2n\pi - \frac{3\pi}{4}, 2n\pi - \frac{\pi}{4} \right) \]
where \( n \) is any integer.