Solve ` |sqrt(3) cos x - sin x |ge 2` for ` x in [0,4pi]` .

Answer

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

### Step 1: Remove the absolute value
We start by rewriting the absolute value inequality:
\[ \sqrt{3} \cos x - \sin x \geq 2 \quad \text{or} \quad \sqrt{3} \cos x - \sin x \leq -2 \]

### Step 2: Solve the first inequality
For the first case:
\[ \sqrt{3} \cos x - \sin x \geq 2 \]
Rearranging gives:
\[ \sqrt{3} \cos x - 2 \geq \sin x \]
This can be rewritten as:
\[ \sqrt{3} \cos x - \sin x \geq 2 \]

### Step 3: Solve the second inequality
For the second case:
\[ \sqrt{3} \cos x - \sin x \leq -2 \]
Rearranging gives:
\[ \sqrt{3} \cos x + 2 \leq \sin x \]
This can be rewritten as:
\[ \sqrt{3} \cos x + 2 \leq \sin x \]

### Step 4: Convert to trigonometric form
We can express \(\sqrt{3} \cos x - \sin x\) in the form \(R \cos(x + \alpha)\). Here, we identify:
- \(R = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\)
- \(\tan \alpha = \frac{-1}{\sqrt{3}} \Rightarrow \alpha = -\frac{\pi}{6}\)

Thus, we can rewrite:
\[ \sqrt{3} \cos x - \sin x = 2 \cos\left(x + \frac{\pi}{6}\right) \]

### Step 5: Set up the inequalities
Now we have:
\[ 2 \cos\left(x + \frac{\pi}{6}\right) \geq 2 \quad \text{or} \quad 2 \cos\left(x + \frac{\pi}{6}\right) \leq -2 \]

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: Solve the inequalities
1. For \(\cos\left(x + \frac{\pi}{6}\right) = 1\):
\[ x + \frac{\pi}{6} = 2k\pi \Rightarrow x = 2k\pi - \frac{\pi}{6} \]
For \(k = 0\), \(x = -\frac{\pi}{6}\) (not in the interval), for \(k = 1\), \(x = \frac{11\pi}{6}\).

2. For \(\cos\left(x + \frac{\pi}{6}\right) = -1\):
\[ x + \frac{\pi}{6} = (2k + 1)\pi \Rightarrow x = (2k + 1)\pi - \frac{\pi}{6} \]
For \(k = 0\), \(x = \frac{5\pi}{6}\), for \(k = 1\), \(x = \frac{17\pi}{6}\), for \(k = 2\), \(x = \frac{23\pi}{6}\).

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

### Final Answer
Thus, the solutions to the inequality \( |\sqrt{3} \cos x - \sin x| \geq 2 \) for \( x \in [0, 4\pi] \) are:
\[ x = \frac{5\pi}{6}, \frac{11\pi}{6}, \frac{17\pi}{6}, \frac{23\pi}{6} \]