Statement I ` x =(kpi)/(13), k in I` does not represent the general solution of trigonometric equation.
` sin 13 x - sin 13 x cos 2x=0`
Statement II Both ` x=r pi , r in I and x=(k pi )/(13) , k in I ` satisfies the trigonometric equation.
`sin 13 x - sin 13 x cos 2x =0`

Answer

To solve the problem, we need to analyze the given trigonometric equation and the statements provided.

### Step 1: Analyze the given equation
The equation we have is:
\[ \sin(13x) - \sin(13x) \cos(2x) = 0 \]

### Step 2: Factor the equation
We can factor out \(\sin(13x)\):
\[ \sin(13x)(1 - \cos(2x)) = 0 \]

### Step 3: Set each factor to zero
This gives us two cases to consider:
1. \(\sin(13x) = 0\)
2. \(1 - \cos(2x) = 0\)

### Step 4: Solve the first case \(\sin(13x) = 0\)
The general solution for \(\sin(\theta) = 0\) is:
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]
Thus,
\[ 13x = n\pi \implies x = \frac{n\pi}{13} \]

### Step 5: Solve the second case \(1 - \cos(2x) = 0\)
This simplifies to:
\[ \cos(2x) = 1 \]
The general solution for \(\cos(\theta) = 1\) is:
\[ \theta = 2m\pi \quad (m \in \mathbb{Z}) \]
Thus,
\[ 2x = 2m\pi \implies x = m\pi \]

### Step 6: Combine the solutions
From the above steps, we have two sets of solutions:
1. \(x = \frac{n\pi}{13}\) where \(n \in \mathbb{Z}\)
2. \(x = m\pi\) where \(m \in \mathbb{Z}\)

### Step 7: Analyze the statements
- **Statement I**: \(x = \frac{k\pi}{13}\) does not represent the general solution.
- **Statement II**: Both \(x = r\pi\) and \(x = \frac{k\pi}{13}\) satisfy the equation.

**Analysis of Statement I**:
The solution \(x = \frac{k\pi}{13}\) is indeed a valid solution of the equation, so Statement I is **false**.

**Analysis of Statement II**:
Both forms of \(x\) are valid solutions to the equation, so Statement II is **true**.

### Conclusion
- Statement I is false.
- Statement II is true.