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 given inequality:
\[ \cos x - \sin x \geq 1 \]
Rearranging this gives:
\[ \cos x \geq 1 + \sin x \]

### Step 2: Graphing the Functions
To find the solution set, we can analyze the graphs of \( \cos x \) and \( 1 + \sin x \).

1. **Graph of \( \cos x \)**:
- The cosine function oscillates between -1 and 1.
- It starts at 1 when \( x = 0 \), decreases to 0 at \( x = \frac{\pi}{2} \), reaches -1 at \( x = \pi \), returns to 0 at \( x = \frac{3\pi}{2} \), and goes back to 1 at \( x = 2\pi \).

2. **Graph of \( 1 + \sin x \)**:
- The sine function oscillates between -1 and 1, so \( 1 + \sin x \) oscillates between 0 and 2.
- It starts at 1 when \( x = 0 \), reaches 2 at \( x = \frac{\pi}{2} \), goes back to 1 at \( x = \pi \), drops to 0 at \( x = \frac{3\pi}{2} \), and returns to 1 at \( x = 2\pi \).

### Step 3: Finding Intersection Points
Next, we need to find where the graph of \( \cos x \) is above the graph of \( 1 + \sin x \). This means we need to find the points where:
\[ \cos x = 1 + \sin x \]

### Step 4: Solving the Equation
To solve \( \cos x = 1 + \sin x \), we can rearrange it:
\[ \cos x - \sin x = 1 \]
Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \), we can express \( \cos x \) in terms of \( \sin x \):
\[ \cos^2 x = 1 - \sin^2 x \]
Substituting \( \cos x = 1 + \sin x \) into this gives:
\[ (1 + \sin x)^2 + \sin^2 x = 1 \]
Expanding and simplifying:
\[ 1 + 2\sin x + \sin^2 x + \sin^2 x = 1 \]
\[ 2\sin^2 x + 2\sin x = 0 \]
Factoring out \( 2\sin x \):
\[ 2\sin x(\sin x + 1) = 0 \]
This gives us:
\[ \sin x = 0 \quad \text{or} \quad \sin x = -1 \]

### Step 5: Finding Values of \( x \)
1. **For \( \sin x = 0 \)**:
- The solutions in the interval \( [0, 2\pi] \) are:
\[ x = 0, \pi, 2\pi \]

2. **For \( \sin x = -1 \)**:
- The solution in the interval \( [0, 2\pi] \) is:
\[ x = \frac{3\pi}{2} \]

### Step 6: Analyzing the Regions
Now we check the intervals:
- Between \( 0 \) and \( \frac{3\pi}{2} \), \( \cos x \) is above \( 1 + \sin x \).
- Between \( \frac{3\pi}{2} \) and \( 2\pi \), \( \cos x \) is also above \( 1 + \sin x \).

### Final Solution Set
Thus, the solution set for \( x \) where \( \cos x - \sin x \geq 1 \) is:
\[ x \in \left[ \frac{3\pi}{2}, 2\pi \right] \]