The number of values of x in `[0, 4 pi]` satisfying the inequation `|sqrt(3)"cos" x - "sin"x|ge2`, is

Options

A. 0
B. 2
C. 4
D. 8

Explanation

To solve the inequality \( |\sqrt{3} \cos x - \sin x| \geq 2 \) for \( x \) in the interval \( [0, 4\pi] \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ |\sqrt{3} \cos x - \sin x| \geq 2 \]
This can be rewritten as two separate inequalities:
\[ \sqrt{3} \cos x - \sin x \geq 2 \quad \text{or} \quad \sqrt{3} \cos x - \sin x \leq -2 \]

### Step 2: Solve the First Inequality
Let's solve the first inequality:
\[ \sqrt{3} \cos x - \sin x \geq 2 \]
Rearranging gives:
\[ \sqrt{3} \cos x \geq \sin x + 2 \]
Dividing through by 2 gives:
\[ \frac{\sqrt{3}}{2} \cos x \geq \frac{1}{2} \sin x + 1 \]

### Step 3: Solve the Second Inequality
Now, let's solve the second inequality:
\[ \sqrt{3} \cos x - \sin x \leq -2 \]
Rearranging gives:
\[ \sqrt{3} \cos x \leq \sin x - 2 \]
Dividing through by 2 gives:
\[ \frac{\sqrt{3}}{2} \cos x \leq \frac{1}{2} \sin x - 1 \]

### Step 4: Convert to Standard Form
To solve these inequalities, we can express them in terms of a single trigonometric function. We can use the identity:
\[ R \cos(x + \phi) = \sqrt{3} \cos x - \sin x \]
where \( R = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2 \) and \( \phi = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \).

Thus, we can rewrite:
\[ 2 \cos\left(x + \frac{\pi}{6}\right) \geq 2 \quad \text{or} \quad 2 \cos\left(x + \frac{\pi}{6}\right) \leq -2 \]

### Step 5: Simplify the Inequalities
Dividing by 2 gives:
\[ \cos\left(x + \frac{\pi}{6}\right) \geq 1 \quad \text{or} \quad \cos\left(x + \frac{\pi}{6}\right) \leq -1 \]

### Step 6: Analyze the Cosine Function
The inequality \( \cos\left(x + \frac{\pi}{6}\right) = 1 \) occurs at:
\[ x + \frac{\pi}{6} = 2n\pi \implies x = 2n\pi - \frac{\pi}{6} \]
For \( n = 0, 1, 2, 3 \), we find:
- \( n = 0 \): \( x = -\frac{\pi}{6} \) (not in range)
- \( n = 1 \): \( x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \)
- \( n = 2 \): \( x = 4\pi - \frac{\pi}{6} = \frac{23\pi}{6} \)

The inequality \( \cos\left(x + \frac{\pi}{6}\right) = -1 \) occurs at:
\[ x + \frac{\pi}{6} = (2n + 1)\pi \implies x = (2n + 1)\pi - \frac{\pi}{6} \]
For \( n = 0, 1, 2, 3 \), we find:
- \( n = 0 \): \( x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)
- \( n = 1 \): \( x = 3\pi - \frac{\pi}{6} = \frac{17\pi}{6} \)

### Step 7: Collect All Solutions
The valid solutions in the interval \( [0, 4\pi] \) are:
1. \( x = \frac{11\pi}{6} \)
2. \( x = \frac{23\pi}{6} \)
3. \( x = \frac{5\pi}{6} \)
4. \( x = \frac{17\pi}{6} \)

### Conclusion
Thus, the number of values of \( x \) in the interval \( [0, 4\pi] \) satisfying the inequality is:
\[ \boxed{4} \]