Find all the value of ` theta ` satisfying the equation , ` sin 7 theta = sin theta + sin 3 theta ` such that ` 0 le theta le pi`

Answer

To solve the equation \( \sin(7\theta) = \sin(\theta) + \sin(3\theta) \) for \( 0 \leq \theta \leq \pi \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin(7\theta) - \sin(\theta) - \sin(3\theta) = 0 \]
This can be rearranged to:
\[ \sin(7\theta) - \sin(\theta) = \sin(3\theta) \]

### Step 2: Use the sine subtraction formula
Using the sine subtraction formula:
\[ \sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
we can apply it to \( \sin(7\theta) - \sin(\theta) \):
\[ \sin(7\theta) - \sin(\theta) = 2 \cos\left(\frac{7\theta + \theta}{2}\right) \sin\left(\frac{7\theta - \theta}{2}\right) \]
This simplifies to:
\[ 2 \cos(4\theta) \sin(3\theta) \]

### Step 3: Substitute back into the equation
Now we substitute this back into our equation:
\[ 2 \cos(4\theta) \sin(3\theta) = \sin(3\theta) \]

### Step 4: Factor the equation
We can factor out \( \sin(3\theta) \):
\[ \sin(3\theta)(2 \cos(4\theta) - 1) = 0 \]

### Step 5: Solve for \( \sin(3\theta) = 0 \)
The first part of the equation gives us:
\[ \sin(3\theta) = 0 \]
This implies:
\[ 3\theta = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ \theta = \frac{n\pi}{3} \]
Considering the range \( 0 \leq \theta \leq \pi \), the possible values of \( n \) are \( 0, 1, 2, 3 \):
- For \( n = 0 \): \( \theta = 0 \)
- For \( n = 1 \): \( \theta = \frac{\pi}{3} \)
- For \( n = 2 \): \( \theta = \frac{2\pi}{3} \)
- For \( n = 3 \): \( \theta = \pi \)

### Step 6: Solve for \( 2 \cos(4\theta) - 1 = 0 \)
The second part of the equation gives us:
\[ 2 \cos(4\theta) - 1 = 0 \implies \cos(4\theta) = \frac{1}{2} \]
This implies:
\[ 4\theta = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 4\theta = \frac{5\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \]
Thus:
1. From \( 4\theta = \frac{\pi}{3} + 2k\pi \):
\[ \theta = \frac{\pi}{12} + \frac{k\pi}{2} \]
For \( k = 0 \): \( \theta = \frac{\pi}{12} \)
For \( k = 1 \): \( \theta = \frac{7\pi}{12} \)

2. From \( 4\theta = \frac{5\pi}{3} + 2k\pi \):
\[ \theta = \frac{5\pi}{12} + \frac{k\pi}{2} \]
For \( k = 0 \): \( \theta = \frac{5\pi}{12} \)

### Step 7: Collect all solutions
Now we compile all the solutions:
- From \( \sin(3\theta) = 0 \): \( \theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi \)
- From \( \cos(4\theta) = \frac{1}{2} \): \( \theta = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12} \)

### Final Solutions
Thus, the complete set of solutions in the interval \( 0 \leq \theta \leq \pi \) is:
\[ \theta = 0, \frac{\pi}{12}, \frac{\pi}{3}, \frac{5\pi}{12}, \frac{2\pi}{3}, \frac{7\pi}{12}, \pi \]