Solve :` cos 3x + cos 2x = sin ""(3)/(2) x + sin "" (1)/(2) x, 0 lt x le pi.`

Answer

To solve the equation \( \cos(3x) + \cos(2x) = \sin\left(\frac{3}{2} x\right) + \sin\left(\frac{1}{2} x\right) \) for \( 0 < x \leq \pi \), we will follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We can use the sum-to-product identities for both sides of the equation. The identities state:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]

Applying these identities to our equation:
- For the left side:
\[ \cos(3x) + \cos(2x) = 2 \cos\left(\frac{3x + 2x}{2}\right) \cos\left(\frac{3x - 2x}{2}\right) = 2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{x}{2}\right) \]

- For the right side:
\[ \sin\left(\frac{3}{2} x\right) + \sin\left(\frac{1}{2} x\right) = 2 \sin\left(\frac{3x + x}{4}\right) \cos\left(\frac{3x - x}{4}\right) = 2 \sin\left(\frac{2x}{2}\right) \cos\left(\frac{x}{2}\right) = 2 \sin(x) \cos\left(\frac{x}{2}\right) \]

### Step 2: Set the two sides equal
Now we have:
\[ 2 \cos\left(\frac{5x}{2}\right) \cos\left(\frac{x}{2}\right) = 2 \sin(x) \cos\left(\frac{x}{2}\right) \]

### Step 3: Divide both sides by \( 2 \cos\left(\frac{x}{2}\right) \)
Assuming \( \cos\left(\frac{x}{2}\right) \neq 0 \), we can divide both sides by \( 2 \cos\left(\frac{x}{2}\right) \):
\[ \cos\left(\frac{5x}{2}\right) = \sin(x) \]

### Step 4: Use the identity \( \sin(x) = \cos\left(\frac{\pi}{2} - x\right) \)
This gives us:
\[ \cos\left(\frac{5x}{2}\right) = \cos\left(\frac{\pi}{2} - x\right) \]

### Step 5: Set up the equations
From the property of cosine, we have two cases:
1. \( \frac{5x}{2} = \frac{\pi}{2} - x + 2n\pi \)
2. \( \frac{5x}{2} = -\left(\frac{\pi}{2} - x\right) + 2n\pi \)

### Step 6: Solve the first case
For the first case:
\[ \frac{5x}{2} + x = \frac{\pi}{2} + 2n\pi \]
\[ \frac{7x}{2} = \frac{\pi}{2} + 2n\pi \]
\[ x = \frac{\pi + 4n\pi}{7} \]

### Step 7: Solve the second case
For the second case:
\[ \frac{5x}{2} + \frac{\pi}{2} - x = 2n\pi \]
\[ \frac{3x}{2} + \frac{\pi}{2} = 2n\pi \]
\[ \frac{3x}{2} = 2n\pi - \frac{\pi}{2} \]
\[ x = \frac{4n\pi - \pi}{3} = \frac{(4n - 1)\pi}{3} \]

### Step 8: Determine valid solutions for \( n \)
Now we need to find valid values of \( n \) such that \( 0 < x \leq \pi \).

1. From \( x = \frac{\pi + 4n\pi}{7} \):
- For \( n = 0 \): \( x = \frac{\pi}{7} \)
- For \( n = 1 \): \( x = \frac{5\pi}{7} \)
- For \( n = 2 \): \( x = \frac{9\pi}{7} \) (not valid)

2. From \( x = \frac{(4n - 1)\pi}{3} \):
- For \( n = 1 \): \( x = \frac{3\pi}{3} = \pi \)
- For \( n = 2 \): \( x = \frac{7\pi}{3} \) (not valid)

### Final Solutions
The valid solutions are:
\[ x = \frac{\pi}{7}, \quad x = \frac{5\pi}{7}, \quad x = \pi \]