The values of `theta` satisfying `"sin" 7 theta = "sin" 4 theta -"sin" theta " and " 0 lt theta lt (pi)/(2)` are

Options

A. `(pi)/(9), (pi)/(4)`
B. `(pi)/(3), (pi)/(9)`
C. `(pi)/(6), (pi)/(9)`
D. `(pi)/(3), (pi)/(4)`

Answer

To solve the equation \( \sin(7\theta) = \sin(4\theta) - \sin(\theta) \) for \( 0 < \theta < \frac{\pi}{2} \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ \sin(7\theta) = \sin(4\theta) - \sin(\theta) \]
Rearranging gives:
\[ \sin(7\theta) + \sin(\theta) = \sin(4\theta) \]

### Step 2: Use the Sum of Sines Formula
Using the formula for the sum of sines, \( \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \), we can rewrite \( \sin(7\theta) + \sin(\theta) \):
\[ \sin(7\theta) + \sin(\theta) = 2 \sin\left(\frac{7\theta + \theta}{2}\right) \cos\left(\frac{7\theta - \theta}{2}\right) = 2 \sin(4\theta) \cos(3\theta) \]
Thus, we have:
\[ 2 \sin(4\theta) \cos(3\theta) = \sin(4\theta) \]

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

### Step 4: Solve Each Factor
This gives us two equations to solve:
1. \( \sin(4\theta) = 0 \)
2. \( 2\cos(3\theta) - 1 = 0 \)

#### Solving \( \sin(4\theta) = 0 \)
The general solution for \( \sin x = 0 \) is \( x = n\pi \). Therefore:
\[ 4\theta = n\pi \implies \theta = \frac{n\pi}{4} \]
For \( n = 0, 1, 2, \ldots \), we find:
- \( \theta = 0 \)
- \( \theta = \frac{\pi}{4} \)
- \( \theta = \frac{\pi}{2} \)

Since we are looking for \( 0 < \theta < \frac{\pi}{2} \), we have \( \theta = \frac{\pi}{4} \).

#### Solving \( 2\cos(3\theta) - 1 = 0 \)
Rearranging gives:
\[ \cos(3\theta) = \frac{1}{2} \]
The general solutions for \( \cos x = \frac{1}{2} \) are:
\[ 3\theta = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad 3\theta = \frac{5\pi}{3} + 2k\pi \]
Solving for \( \theta \):
1. \( 3\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{9} \)
2. \( 3\theta = \frac{5\pi}{3} \implies \theta = \frac{5\pi}{9} \)

### Step 5: Check Validity of Solutions
Now we check which of these solutions lie within \( 0 < \theta < \frac{\pi}{2} \):
- From \( \sin(4\theta) = 0 \), we have \( \theta = \frac{\pi}{4} \).
- From \( 2\cos(3\theta) - 1 = 0 \), we have \( \theta = \frac{\pi}{9} \) and \( \theta = \frac{5\pi}{9} \) (but \( \frac{5\pi}{9} > \frac{\pi}{2} \)).

Thus, the valid solutions are:
\[ \theta = \frac{\pi}{4} \quad \text{and} \quad \theta = \frac{\pi}{9} \]

### Final Answer
The values of \( \theta \) satisfying the equation in the interval \( 0 < \theta < \frac{\pi}{2} \) are:
\[ \theta = \frac{\pi}{4}, \frac{\pi}{9} \]