If ` 2 cos x lt sqrt(3) and x in [-pi, pi]` , then find the solution set for x .

Answer

To solve the inequality \( 2 \cos x < \sqrt{3} \) for \( x \) in the interval \([-π, π]\), we can follow these steps:

### Step 1: Simplify the Inequality
We start with the inequality:
\[ 2 \cos x < \sqrt{3} \]
Dividing both sides by 2 gives:
\[ \cos x < \frac{\sqrt{3}}{2} \]

### Step 2: Identify Critical Points
The cosine function equals \( \frac{\sqrt{3}}{2} \) at specific angles. We know:
\[ \cos x = \frac{\sqrt{3}}{2} \quad \text{at} \quad x = \frac{\pi}{6} \quad \text{and} \quad x = -\frac{\pi}{6} \]
These points divide the interval \([-π, π]\) into segments.

### Step 3: Determine Intervals
We will analyze the sign of \( \cos x - \frac{\sqrt{3}}{2} \) in the intervals:
1. \( (-π, -\frac{\pi}{6}) \)
2. \( (-\frac{\pi}{6}, \frac{\pi}{6}) \)
3. \( (\frac{\pi}{6}, π) \)

### Step 4: Test Each Interval
1. **Interval \( (-π, -\frac{\pi}{6}) \)**:
- Choose \( x = -\frac{\pi}{4} \):
\[ \cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \quad (\text{which is less than } \frac{\sqrt{3}}{2}) \]
Thus, \( \cos x < \frac{\sqrt{3}}{2} \) is true.

2. **Interval \( (-\frac{\pi}{6}, \frac{\pi}{6}) \)**:
- Choose \( x = 0 \):
\[ \cos(0) = 1 \quad (\text{which is greater than } \frac{\sqrt{3}}{2}) \]
Thus, \( \cos x < \frac{\sqrt{3}}{2} \) is false.

3. **Interval \( (\frac{\pi}{6}, π) \)**:
- Choose \( x = \frac{\pi}{2} \):
\[ \cos(\frac{\pi}{2}) = 0 \quad (\text{which is less than } \frac{\sqrt{3}}{2}) \]
Thus, \( \cos x < \frac{\sqrt{3}}{2} \) is true.

### Step 5: Combine the Results
From the tests, we find that the solution set where \( 2 \cos x < \sqrt{3} \) is:
\[ x \in \left[-π, -\frac{\pi}{6}\right) \cup \left(\frac{\pi}{6}, π\right] \]

### Final Solution
Thus, the solution set for \( x \) is:
\[ \boxed{[-π, -\frac{\pi}{6}) \cup (\frac{\pi}{6}, π]} \]