Solve ` |sin 3x + sinx | + |sin 3x - sinx |= sqrt3, - (pi)/(2) lt x lt (pi)/(2).`

Answer

To solve the equation \( | \sin 3x + \sin x | + | \sin 3x - \sin x | = \sqrt{3} \) for \( -\frac{\pi}{2} < x < \frac{\pi}{2} \), we will break it down into manageable steps.

### Step 1: Analyze the Absolute Values
The expression involves absolute values, which means we need to consider different cases based on the signs of \( \sin 3x + \sin x \) and \( \sin 3x - \sin x \).

### Step 2: Case Analysis
1. **Case 1:** \( \sin 3x + \sin x \geq 0 \) and \( \sin 3x - \sin x \geq 0 \)
- Here, the equation simplifies to:
\[ \sin 3x + \sin x + \sin 3x - \sin x = \sqrt{3} \]
\[ 2 \sin 3x = \sqrt{3} \]
\[ \sin 3x = \frac{\sqrt{3}}{2} \]
- The solutions for \( 3x \) are:
\[ 3x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 3x = \frac{2\pi}{3} + 2k\pi \]
- Thus,
\[ x = \frac{\pi}{9} + \frac{2k\pi}{3} \quad \text{or} \quad x = \frac{2\pi}{9} + \frac{2k\pi}{3} \]
- For \( k = 0 \):
\[ x = \frac{\pi}{9}, \frac{2\pi}{9} \]

2. **Case 2:** \( \sin 3x + \sin x \geq 0 \) and \( \sin 3x - \sin x < 0 \)
- The equation simplifies to:
\[ \sin 3x + \sin x - (\sin 3x - \sin x) = \sqrt{3} \]
\[ 2 \sin x = \sqrt{3} \]
\[ \sin x = \frac{\sqrt{3}}{2} \]
- The solution for \( x \) is:
\[ x = \frac{\pi}{3} + 2k\pi \]
- For \( k = 0 \):
\[ x = \frac{\pi}{3} \]

3. **Case 3:** \( \sin 3x + \sin x < 0 \) and \( \sin 3x - \sin x < 0 \)
- The equation simplifies to:
\[ -(\sin 3x + \sin x) - (\sin 3x - \sin x) = \sqrt{3} \]
\[ -2 \sin 3x = \sqrt{3} \]
\[ \sin 3x = -\frac{\sqrt{3}}{2} \]
- The solutions for \( 3x \) are:
\[ 3x = -\frac{\pi}{3} + 2k\pi \quad \text{or} \quad 3x = -\frac{2\pi}{3} + 2k\pi \]
- Thus,
\[ x = -\frac{\pi}{9} + \frac{2k\pi}{3} \quad \text{or} \quad x = -\frac{2\pi}{9} + \frac{2k\pi}{3} \]
- For \( k = 0 \):
\[ x = -\frac{\pi}{9}, -\frac{2\pi}{9} \]

### Step 3: Compile All Solutions
From the cases analyzed, we have the following potential solutions:
- From Case 1: \( x = \frac{\pi}{9}, \frac{2\pi}{9} \)
- From Case 2: \( x = \frac{\pi}{3} \)
- From Case 3: \( x = -\frac{\pi}{9}, -\frac{2\pi}{9} \)

### Step 4: Check Validity Within the Interval
We need to ensure all solutions are within the interval \( -\frac{\pi}{2} < x < \frac{\pi}{2} \):
- \( \frac{\pi}{9} \) is valid.
- \( \frac{2\pi}{9} \) is valid.
- \( \frac{\pi}{3} \) is valid.
- \( -\frac{\pi}{9} \) is valid.
- \( -\frac{2\pi}{9} \) is valid.

### Final Solutions
Thus, the final solutions are:
\[ x = -\frac{2\pi}{9}, -\frac{\pi}{9}, \frac{\pi}{9}, \frac{2\pi}{9}, \frac{\pi}{3} \]