The number of values of `x` in `[0,2pi]` satisfying the equation `|cos x – sin x| >= sqrt2` is

Options

A. 0
B. 1
C. 2
D. none of these

Explanation

To solve the equation \( | \cos x - \sin x | \geq \sqrt{2} \) for \( x \) in the interval \( [0, 2\pi] \), we will follow these steps:

### Step 1: Rewrite the Absolute Value Inequality
The absolute value inequality can be split into two cases:
1. \( \cos x - \sin x \geq \sqrt{2} \)
2. \( \cos x - \sin x \leq -\sqrt{2} \)

### Step 2: Solve the First Case
For the first case, \( \cos x - \sin x \geq \sqrt{2} \):
\[ \cos x - \sin x \geq \sqrt{2} \]
This can be rewritten as:
\[ \cos x \geq \sin x + \sqrt{2} \]
To analyze this, we can express \( \cos x - \sin x \) in terms of a single trigonometric function. We know that:
\[ \cos x - \sin x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x \right) \]
This can be rewritten using the angle subtraction formula:
\[ \cos x - \sin x = \sqrt{2} \cos\left(x + \frac{\pi}{4}\right) \]
Thus, we have:
\[ \sqrt{2} \cos\left(x + \frac{\pi}{4}\right) \geq \sqrt{2} \]
Dividing both sides by \( \sqrt{2} \):
\[ \cos\left(x + \frac{\pi}{4}\right) \geq 1 \]
The only solution for \( \cos\theta = 1 \) is:
\[ x + \frac{\pi}{4} = 2k\pi \quad \text{for integer } k \]
This gives:
\[ x = 2k\pi - \frac{\pi}{4} \]
For \( k = 0 \):
\[ x = -\frac{\pi}{4} \quad \text{(not in } [0, 2\pi]) \]
For \( k = 1 \):
\[ x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \quad \text{(valid)} \]
Thus, the first case gives us one solution: \( x = \frac{7\pi}{4} \).

### Step 3: Solve the Second Case
For the second case, \( \cos x - \sin x \leq -\sqrt{2} \):
\[ \cos x - \sin x \leq -\sqrt{2} \]
This can be rewritten as:
\[ \cos x \leq \sin x - \sqrt{2} \]
Using the same transformation:
\[ \sqrt{2} \cos\left(x + \frac{\pi}{4}\right) \leq -\sqrt{2} \]
Dividing both sides by \( \sqrt{2} \):
\[ \cos\left(x + \frac{\pi}{4}\right) \leq -1 \]
The only solution for \( \cos\theta = -1 \) is:
\[ x + \frac{\pi}{4} = (2k + 1)\pi \quad \text{for integer } k \]
This gives:
\[ x = (2k + 1)\pi - \frac{\pi}{4} \]
For \( k = 0 \):
\[ x = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \quad \text{(valid)} \]
For \( k = 1 \):
\[ x = 3\pi - \frac{\pi}{4} = \frac{11\pi}{4} \quad \text{(not in } [0, 2\pi]) \]
Thus, the second case gives us one solution: \( x = \frac{3\pi}{4} \).

### Step 4: Combine Solutions
From both cases, we have found two solutions:
1. \( x = \frac{7\pi}{4} \)
2. \( x = \frac{3\pi}{4} \)

### Conclusion
The total number of values of \( x \) in the interval \( [0, 2\pi] \) satisfying the equation \( | \cos x - \sin x | \geq \sqrt{2} \) is **2**.

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