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 \( x \in [0, 4\pi] \), 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 split into 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
For 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:
\[ \frac{\sqrt{3}}{2} \cos x \geq \frac{1}{2} \sin x + 1 \]
This can be rewritten using the sine addition formula:
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \geq 1 \]
This is equivalent to:
\[ \sin\left(\frac{\pi}{3} - x\right) \geq 1 \]
The sine function achieves a maximum value of 1, so we set:
\[ \frac{\pi}{3} - x = \frac{\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]
Solving for \( x \):
\[ x = \frac{\pi}{3} - \frac{\pi}{2} - 2k\pi = -\frac{\pi}{6} - 2k\pi \]
This solution does not yield valid values in the interval \( [0, 4\pi] \).

### Step 3: Solve the second inequality
Now we 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:
\[ \frac{\sqrt{3}}{2} \cos x \leq \frac{1}{2} \sin x - 1 \]
This can be rewritten using the sine addition formula:
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x \leq -1 \]
This is equivalent to:
\[ \sin\left(\frac{\pi}{3} - x\right) \leq -1 \]
The sine function achieves a minimum value of -1, so we set:
\[ \frac{\pi}{3} - x = \frac{3\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]
Solving for \( x \):
\[ x = \frac{\pi}{3} - \frac{3\pi}{2} - 2k\pi = -\frac{7\pi}{6} - 2k\pi \]
Again, this solution does not yield valid values in the interval \( [0, 4\pi] \).

### Step 4: Finding valid solutions
We can also check for specific angles where the sine function equals 1 or -1:
1. \( \frac{\pi}{3} - x = \frac{\pi}{2} \) gives \( x = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6} \) (not valid).
2. \( \frac{\pi}{3} - x = \frac{3\pi}{2} \) gives \( x = \frac{\pi}{3} - \frac{3\pi}{2} = -\frac{7\pi}{6} \) (not valid).
3. Continue checking for valid values of \( k \) to find solutions within \( [0, 4\pi] \).

### Final Solutions
After checking for valid \( k \) values, we find:
1. \( x = \frac{5\pi}{6} \)
2. \( x = \frac{11\pi}{6} \)
3. \( x = \frac{17\pi}{6} \)
4. \( x = \frac{23\pi}{6} \)

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