If `sin5x+sin3x+sinx=0`, then the value of x other than zero between `0lexlepi//2` is

Options

A. `pi//6`
B. `pi//12`
C. `pi//3`
D. `pi//9`

Answer

To solve the equation \( \sin 5x + \sin 3x + \sin x = 0 \) for values of \( x \) other than zero in the interval \( \left(0, \frac{\pi}{2}\right) \), we can follow these steps:

### Step 1: Use the sum-to-product identities
We can group the terms and apply the sum-to-product identities. The identity states that:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Let's group \( \sin 5x \) and \( \sin x \):
\[ \sin 5x + \sin x = 2 \sin\left(\frac{5x + x}{2}\right) \cos\left(\frac{5x - x}{2}\right) = 2 \sin(3x) \cos(2x) \]
Thus, we can rewrite the equation as:
\[ 2 \sin(3x) \cos(2x) + \sin 3x = 0 \]

### Step 2: Factor out \( \sin 3x \)
Now, we can factor out \( \sin 3x \):
\[ \sin 3x (2 \cos 2x + 1) = 0 \]

### Step 3: Solve for \( \sin 3x = 0 \)
Setting \( \sin 3x = 0 \) gives:
\[ 3x = n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ x = \frac{n\pi}{3} \]

### Step 4: Solve for \( 2 \cos 2x + 1 = 0 \)
Setting \( 2 \cos 2x + 1 = 0 \) gives:
\[ \cos 2x = -\frac{1}{2} \]
The angles where cosine is \(-\frac{1}{2}\) are:
\[ 2x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad 2x = \frac{4\pi}{3} + 2k\pi \quad \text{for } k \in \mathbb{Z} \]
Thus,
\[ x = \frac{\pi}{3} + k\pi \quad \text{or} \quad x = \frac{2\pi}{3} + k\pi \]

### Step 5: Find values of \( x \) in the interval \( (0, \frac{\pi}{2}) \)
Now we need to find values of \( x \) from both cases that lie in the interval \( (0, \frac{\pi}{2}) \).

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

2. From \( x = \frac{\pi}{3} + k\pi \):
- For \( k = 0 \): \( x = \frac{\pi}{3} \) (valid)
- For \( k = 1 \): \( x = \frac{4\pi}{3} \) (not valid, exceeds \( \frac{\pi}{2} \))

3. From \( x = \frac{2\pi}{3} + k\pi \):
- For \( k = 0 \): \( x = \frac{2\pi}{3} \) (not valid, exceeds \( \frac{\pi}{2} \))

### Conclusion
The only valid solution in the interval \( (0, \frac{\pi}{2}) \) is:
\[ \boxed{\frac{\pi}{3}} \]