If cosx−sinx≥1 and 0≤x≤2π, then the solution set for x is __

Options

A. `[0, pi//4] cup [7 pi//4, 2pi]`
B. `[3 pi//2, 7 pi//4] cup {0}`
C. `[3 pi//2, 2pi] cup {0}`
D. none of these

Explanation

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

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

### Step 2: Rearrange the inequality
Rearranging gives:
\[ \cos x - 1 \geq \sin x \]

### Step 3: Square both sides
To eliminate the sine term, we can square both sides. However, we need to be cautious about the conditions under which this is valid. Let's first isolate the terms:
\[ \cos x - 1 \geq \sin x \implies (\cos x - 1)^2 \geq \sin^2 x \]

### Step 4: Use the Pythagorean identity
Recall that \( \sin^2 x + \cos^2 x = 1 \). Thus, we can express \( \sin^2 x \) in terms of \( \cos^2 x \):
\[ (\cos x - 1)^2 \geq 1 - \cos^2 x \]

### Step 5: Expand and simplify
Expanding the left side:
\[ \cos^2 x - 2\cos x + 1 \geq 1 - \cos^2 x \]
This simplifies to:
\[ \cos^2 x - 2\cos x + 1 + \cos^2 x \geq 0 \]
\[ 2\cos^2 x - 2\cos x \geq 0 \]

### Step 6: Factor the inequality
Factoring out \( 2\cos x \):
\[ 2\cos x(\cos x - 1) \geq 0 \]
This gives us two factors to consider:
1. \( 2\cos x \geq 0 \)
2. \( \cos x - 1 \geq 0 \)

### Step 7: Solve each factor
1. From \( 2\cos x \geq 0 \):
\[ \cos x \geq 0 \implies x \in [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi] \]

2. From \( \cos x - 1 \geq 0 \):
\[ \cos x = 1 \implies x = 0 \]

### Step 8: Combine the solutions
The solution set must satisfy both conditions. The only value that satisfies both is:
\[ x = 0 \]

### Final Solution
Thus, the solution set for \( \cos x - \sin x \geq 1 \) in the interval \( [0, 2\pi] \) is:
\[ \{0\} \]