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

Options

A. `(pi)/(6)`
B. `(pi)/(12)`
C. `(pi)/(3)`
D. `(pi)/(4)`

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin 5x + \sin 3x + \sin x = 0 \]

### Step 2: Group the terms
We can group the first two sine terms:
\[ \sin 5x + \sin 3x = -\sin x \]

### Step 3: Use the sine addition formula
Using the formula for the sum of sines:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
where \( A = 5x \) and \( B = 3x \), we have:
\[ \sin 5x + \sin 3x = 2 \sin\left(\frac{5x + 3x}{2}\right) \cos\left(\frac{5x - 3x}{2}\right) = 2 \sin(4x) \cos(x) \]
Thus, we can rewrite our equation as:
\[ 2 \sin(4x) \cos(x) + \sin x = 0 \]

### Step 4: Factor the equation
Now, we can factor out \( \sin x \):
\[ \sin x (2 \sin(4x) \cos(x) + 1) = 0 \]
This gives us two cases to consider:
1. \( \sin x = 0 \)
2. \( 2 \sin(4x) \cos(x) + 1 = 0 \)

### Step 5: Solve the first case
For \( \sin x = 0 \):
\[ x = n\pi \quad (n \in \mathbb{Z}) \]
In the interval \( 0 \leq x \leq \frac{\pi}{2} \), the only solution is \( x = 0 \).

### Step 6: Solve the second case
For \( 2 \sin(4x) \cos(x) + 1 = 0 \):
\[ 2 \sin(4x) \cos(x) = -1 \]
\[ \sin(4x) \cos(x) = -\frac{1}{2} \]

### Step 7: Analyze the equation
Since \( \sin(4x) \) and \( \cos(x) \) are both non-negative in the interval \( 0 \leq x \leq \frac{\pi}{2} \), the left side cannot be negative. Thus, we need to find when:
\[ \sin(4x) = -\frac{1}{2} \]
However, since \( \sin(4x) \) is non-negative in this interval, we discard this case.

### Step 8: Find the valid solution
Returning to our earlier results, the only valid solution in the interval \( 0 \leq x \leq \frac{\pi}{2} \) other than \( x = 0 \) is:
\[ x = \frac{\pi}{3} \]

### Conclusion
Thus, the value of \( x \) other than 0 lying between \( 0 \leq x \leq \frac{\pi}{2} \) is:
\[ \boxed{\frac{\pi}{3}} \]