If ` cos x- sin x ge 1 and 0 le x le 2pi` , then find the solution set for x .

Answer

To solve the inequality \( \cos x - \sin x \geq 1 \) for \( 0 \leq x \leq 2\pi \), we will follow these steps:

### Step 1: Rearranging the Inequality
We start with the inequality:
\[ \cos x - \sin x \geq 1 \]
Rearranging gives:
\[ \cos x - 1 \geq \sin x \]

### Step 2: Multiplying by \( \frac{1}{\sqrt{2}} \)
Next, we multiply both sides of the inequality by \( \frac{1}{\sqrt{2}} \):
\[ \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x \geq \frac{1}{\sqrt{2}} \]
This can be rewritten using the angle \( \frac{\pi}{4} \):
\[ \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x = \cos\left(x + \frac{\pi}{4}\right) \]
Thus, we have:
\[ \cos\left(x + \frac{\pi}{4}\right) \geq \frac{1}{\sqrt{2}} \]

### Step 3: Finding the General Solution
The inequality \( \cos A \geq \frac{1}{\sqrt{2}} \) holds for:
\[ A \in [0, \frac{\pi}{4}] \cup [\frac{7\pi}{4}, 2\pi] \]
Substituting back for \( A \):
\[ x + \frac{\pi}{4} \in [0, \frac{\pi}{4}] \cup [\frac{7\pi}{4}, 2\pi] \]

### Step 4: Solving for \( x \)
1. From the first interval:
\[ 0 \leq x + \frac{\pi}{4} \leq \frac{\pi}{4} \]
Subtracting \( \frac{\pi}{4} \):
\[ -\frac{\pi}{4} \leq x \leq 0 \]
Since \( x \) must be in the range \( [0, 2\pi] \), the only valid solution here is:
\[ x = 0 \]

2. From the second interval:
\[ \frac{7\pi}{4} \leq x + \frac{\pi}{4} \leq 2\pi \]
Subtracting \( \frac{\pi}{4} \):
\[ \frac{7\pi}{4} - \frac{\pi}{4} \leq x \leq 2\pi - \frac{\pi}{4} \]
Simplifying gives:
\[ \frac{6\pi}{4} \leq x \leq \frac{8\pi}{4} - \frac{\pi}{4} \]
Which simplifies to:
\[ \frac{3\pi}{2} \leq x \leq \frac{7\pi}{4} \]

### Step 5: Final Solution Set
Combining the results from both intervals, we find:
\[ x = 0 \quad \text{and} \quad \frac{3\pi}{2} \leq x \leq \frac{7\pi}{4} \]

Thus, the solution set is:
\[ \{0\} \cup \left[\frac{3\pi}{2}, \frac{7\pi}{4}\right] \]