Find the number of solutions for the equation `sin 5x+sin 3x+sin x=0` for `0 le x le pi`.

Explanation

To solve the equation \( \sin 5x + \sin 3x + \sin x = 0 \) for \( 0 \leq x \leq \pi \), we can follow these steps:

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

### Step 2: Use the sine addition formula
We can group the first two sine terms:
\[ \sin 5x + \sin 3x = 2 \sin\left(\frac{5x + 3x}{2}\right) \cos\left(\frac{5x - 3x}{2}\right) = 2 \sin(4x) \cos(x) \]
Thus, we can rewrite the equation as:
\[ 2 \sin(4x) \cos(x) + \sin x = 0 \]

### Step 3: Factor the equation
Now, we can factor out \( \sin x \):
\[ \sin x (2 \sin(4x) \cos(x) + 1) = 0 \]

### Step 4: Set each factor to zero
We have two cases to consider:
1. \( \sin x = 0 \)
2. \( 2 \sin(4x) \cos(x) + 1 = 0 \)

### Step 5: Solve \( \sin x = 0 \)
For \( \sin x = 0 \) in the interval \( [0, \pi] \):
\[ x = 0 \quad \text{and} \quad x = \pi \]
So, we have two solutions from this case.

### Step 6: Solve \( 2 \sin(4x) \cos(x) + 1 = 0 \)
Rearranging gives:
\[ 2 \sin(4x) \cos(x) = -1 \]
This leads to:
\[ \sin(4x) \cos(x) = -\frac{1}{2} \]

### Step 7: Analyze the equation \( \sin(4x) \cos(x) = -\frac{1}{2} \)
This equation is more complex, so we will analyze it graphically or numerically.

1. The maximum value of \( \sin(4x) \) is 1, and the minimum value is -1.
2. The maximum value of \( \cos(x) \) is 1, and the minimum value is -1.

Thus, the product \( \sin(4x) \cos(x) \) can take values between -1 and 1.

### Step 8: Find solutions for \( \sin(4x) = -\frac{1}{2\cos(x)} \)
To find the values of \( x \) that satisfy this equation, we can analyze the behavior of \( \sin(4x) \) and \( \cos(x) \) in the interval \( [0, \pi] \).

### Step 9: Count the number of solutions
From the analysis:
- From \( \sin x = 0 \), we have \( x = 0 \) and \( x = \pi \) (2 solutions).
- For \( 2 \sin(4x) \cos(x) + 1 = 0 \), we need to check how many times \( \sin(4x) \) intersects the line \( -\frac{1}{2\cos(x)} \) in the interval \( [0, \pi] \).

### Conclusion
After analyzing both cases, we find that:
- The solutions from \( \sin x = 0 \) contribute 2 solutions.
- The solutions from \( 2 \sin(4x) \cos(x) + 1 = 0 \) need to be evaluated graphically or numerically, but typically, this will yield additional solutions.

Thus, the total number of solutions for the equation \( \sin 5x + \sin 3x + \sin x = 0 \) in the interval \( [0, \pi] \) is **4**.