Number of solutions (s) of the equation `(sin x)/(cos 3x) +(sin 3x)/(cos 9x)+(sin 9x)/(cos 27 x)=0` in the interval `(0, pi/4)` is __________.

Explanation

To solve the equation

\[ \frac{\sin x}{\cos 3x} + \frac{\sin 3x}{\cos 9x} + \frac{\sin 9x}{\cos 27x} = 0 \]

in the interval \( (0, \frac{\pi}{4}) \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:

\[ \frac{\sin x}{\cos 3x} + \frac{\sin 3x}{\cos 9x} + \frac{\sin 9x}{\cos 27x} = 0 \]

### Step 2: Multiply by Cosine Terms
To eliminate the denominators, we can multiply through by \(\cos 3x \cdot \cos 9x \cdot \cos 27x\):

\[ \sin x \cdot \cos 9x \cdot \cos 27x + \sin 3x \cdot \cos 3x \cdot \cos 27x + \sin 9x \cdot \cos 3x \cdot \cos 9x = 0 \]

### Step 3: Use Trigonometric Identities
Using the identity \(2 \sin A \cos A = \sin 2A\), we can rewrite the terms:

- \(\sin x \cdot \cos 9x \cdot \cos 27x\) can be expressed as \(\frac{1}{2} \sin 2x \cdot \cos 9x \cdot \cos 27x\)
- \(\sin 3x \cdot \cos 3x \cdot \cos 27x\) can be expressed as \(\frac{1}{2} \sin 6x \cdot \cos 27x\)
- \(\sin 9x \cdot \cos 3x \cdot \cos 9x\) can be expressed as \(\frac{1}{2} \sin 18x\)

Thus, the equation simplifies to:

\[ \frac{1}{2} \left( \sin 2x \cdot \cos 9x \cdot \cos 27x + \sin 6x \cdot \cos 27x + \sin 18x \right) = 0 \]

### Step 4: Set Each Term to Zero
For the equation to hold, we need:

\[ \sin 2x \cdot \cos 9x \cdot \cos 27x + \sin 6x \cdot \cos 27x + \sin 18x = 0 \]

### Step 5: Analyze Each Term
We can analyze the zeros of each sine function:

1. \(\sin 2x = 0\) gives \(2x = n\pi \Rightarrow x = \frac{n\pi}{2}\)
2. \(\sin 6x = 0\) gives \(6x = m\pi \Rightarrow x = \frac{m\pi}{6}\)
3. \(\sin 18x = 0\) gives \(18x = k\pi \Rightarrow x = \frac{k\pi}{18}\)

### Step 6: Find Solutions in the Interval
Now we need to find the values of \(n\), \(m\), and \(k\) such that \(x\) lies in the interval \( (0, \frac{\pi}{4}) \):

- For \(x = \frac{n\pi}{2}\): The only valid \(n\) is \(n=0\) (gives \(x=0\)), which is not in the interval.
- For \(x = \frac{m\pi}{6}\): Valid values are \(m=1, 2, 3, 4\) (gives \(x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}\)), only \(m=1\) is valid (gives \(x=\frac{\pi}{6}\)).
- For \(x = \frac{k\pi}{18}\): Valid values are \(k=1, 2, 3, 4, 5, 6\) (gives \(x = \frac{\pi}{18}, \frac{2\pi}{18}, \frac{3\pi}{18}, \frac{4\pi}{18}, \frac{5\pi}{18}, \frac{6\pi}{18}\)), which gives \(k=1, 2, 3, 4, 5, 6\) (gives \(x=\frac{\pi}{18}, \frac{\pi}{9}, \frac{5\pi}{18}, \frac{2\pi}{9}, \frac{5\pi}{18}, \frac{\pi/3}\)).

### Step 7: Count the Valid Solutions
The valid solutions in the interval \( (0, \frac{\pi}{4}) \) are:

1. \(x = \frac{\pi}{18}\)
2. \(x = \frac{2\pi}{18} = \frac{\pi}{9}\)
3. \(x = \frac{3\pi}{18} = \frac{\pi}{6}\)
4. \(x = \frac{4\pi}{18} = \frac{2\pi}{9}\)
5. \(x = \frac{5\pi}{18}\)
6. \(x = \frac{6\pi}{18} = \frac{\pi}{3}\)

Thus, the total number of solutions in the interval \( (0, \frac{\pi}{4}) \) is **6**.

### Final Answer
The number of solutions \(s\) of the equation in the interval \( (0, \frac{\pi}{4}) \) is **6**.