Step 1: Understanding the Concept:
This problem involves solving a trigonometric equation. The key is to use the sum-to-product trigonometric identities to simplify the equation and find the possible values for x.
Step 2: Key Formula or Approach:
We will use the sum-to-product formula:
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
We will group $\sin 5x$ and $\sin x$ together to apply this formula.
Step 3: Detailed Explanation:
The given equation is:
\[ \sin 5x + \sin 3x + \sin x = 0 \]
Rearrange the terms:
\[ (\sin 5x + \sin x) + \sin 3x = 0 \]
Apply the sum-to-product formula with $A=5x$ and $B=x$:
\[ 2 \sin\left(\frac{5x+x}{2}\right) \cos\left(\frac{5x-x}{2}\right) + \sin 3x = 0 \]
\[ 2 \sin(3x) \cos(2x) + \sin 3x = 0 \]
Factor out the common term $\sin 3x$:
\[ \sin 3x (2 \cos 2x + 1) = 0 \]
This equation is true if either $\sin 3x = 0$ or $2 \cos 2x + 1 = 0$.
Case 1: $\sin 3x = 0$
The general solution for $\sin \theta = 0$ is $\theta = n\pi$, where n is an integer.
So, $3x = n\pi \implies x = \frac{n\pi}{3}$.
For $n=0$, $x=0$ (which we need to exclude as per the question "other than zero").
For $n=1$, $x = \frac{\pi}{3}$. This value lies in the given interval $0 \le x \le \frac{\pi}{2}$.
Case 2: $2 \cos 2x + 1 = 0$
\[ \cos 2x = -\frac{1}{2} \]
The principal values for $2x$ where cosine is $-1/2$ are $2x = \frac{2\pi}{3}$ and $2x = \frac{4\pi}{3}$.
If $2x = \frac{2\pi}{3}$, then $x = \frac{\pi}{3}$. This is the same solution as in Case 1.
If $2x = \frac{4\pi}{3}$, then $x = \frac{2\pi}{3}$. This value is outside the interval $0 \le x \le \frac{\pi}{2}$.
Step 4: Final Answer:
The only non-zero solution in the given interval is $x = \frac{\pi}{3}$. Therefore, option (B) is correct.