The solution set of the equation
`("sin" 13^(@))^("cot" 3x+ "cot"x) = "sin"^(2) (4pi-x)-"cos" (3pi-x) "cos" (2 pi +x)` is given by

Options

A. `x = (n pi)/(4), n ne 4 lambda, lambda in Z`
B. `x = (n pi)/(2), n ne 2 lambda, lambda in Z`
C. `x = (n pi)/(3), n in 3 lambda`
D. none of these

Explanation

To solve the equation \( (\sin 13^\circ)^{(\cot 3x + \cot x)} = \sin^2(4\pi - x) - \cos(3\pi - x) \cos(2\pi + x) \), we will follow these steps:

### Step 1: Simplify the Right-Hand Side
First, we simplify the right-hand side of the equation:
\[ \sin^2(4\pi - x) - \cos(3\pi - x) \cos(2\pi + x) \]
Using the properties of sine and cosine:
- \(\sin(4\pi - x) = -\sin(x)\) (since \(4\pi\) is a full rotation)
- \(\cos(3\pi - x) = -\cos(x)\)
- \(\cos(2\pi + x) = \cos(x)\)

Thus, we have:
\[ \sin^2(4\pi - x) = \sin^2(x) \]
And,
\[ \cos(3\pi - x) \cos(2\pi + x) = (-\cos(x))(\cos(x)) = -\cos^2(x) \]
So the right-hand side becomes:
\[ \sin^2(x) + \cos^2(x) = 1 \]

### Step 2: Set the Equation
Now, we can rewrite our equation as:
\[ (\sin 13^\circ)^{(\cot 3x + \cot x)} = 1 \]

### Step 3: Analyze the Equation
The expression \(a^b = 1\) holds true if:
1. \(a = 1\) (which is not the case here since \(\sin 13^\circ \neq 1\))
2. \(b = 0\) (which leads us to the next step)
3. \(a = -1\) and \(b\) is an even integer (not applicable here as \(\sin 13^\circ\) is positive)

Thus, we set:
\[ \cot 3x + \cot x = 0 \]

### Step 4: Solve for \(x\)
We rewrite the equation:
\[ \cot 3x = -\cot x \]
This implies:
\[ \cot 3x = \frac{\cos 3x}{\sin 3x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x} \]
So:
\[ \frac{\cos 3x}{\sin 3x} + \frac{\cos x}{\sin x} = 0 \]
Combining these fractions gives:
\[ \frac{\cos 3x \sin x + \cos x \sin 3x}{\sin 3x \sin x} = 0 \]
This leads to:
\[ \cos 3x \sin x + \cos x \sin 3x = 0 \]
Using the sine addition formula:
\[ \sin(3x + x) = 0 \implies \sin(4x) = 0 \]

### Step 5: Solve for \(x\)
The general solution for \(\sin(4x) = 0\) is:
\[ 4x = n\pi \quad \text{where } n \in \mathbb{Z} \]
Thus:
\[ x = \frac{n\pi}{4} \]

### Step 6: Exclude Undefined Values
Since \(\cot 3x\) and \(\cot x\) are undefined for \(x = n\pi\), we exclude these values. Therefore, the solution set is:
\[ x = \frac{n\pi}{4} \quad \text{where } n \text{ is not a multiple of } 4. \]

### Final Answer
The solution set of the given equation is:
\[ x = \frac{n\pi}{4}, \quad n \in \mathbb{Z} \text{ and } n \text{ is not a multiple of } 4. \]