If the number of principal solutions of the equation `tan(7pi cos x)=cot(7pi sin x)` is k, then `(k)/(5)` is equal to

Answer

To solve the equation \( \tan(7\pi \cos x) = \cot(7\pi \sin x \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ \tan(7\pi \cos x) = \cot(7\pi \sin x) \]
Using the identity \( \cot \theta = \tan\left(\frac{\pi}{2} - \theta\), we can rewrite the equation as:
\[ \tan(7\pi \cos x) = \tan\left(\frac{\pi}{2} - 7\pi \sin x\right) \]

### Step 2: Set Up the General Solution
From the property of the tangent function, we know that if \( \tan A = \tan B \), then:
\[ A = n\pi + B \quad \text{for } n \in \mathbb{Z} \]
Applying this to our equation gives:
\[ 7\pi \cos x = n\pi + \left(\frac{\pi}{2} - 7\pi \sin x\right) \]

### Step 3: Simplify the Equation
Rearranging the equation, we have:
\[ 7\pi \cos x + 7\pi \sin x = n\pi + \frac{\pi}{2} \]
Factoring out \( \pi \):
\[ 7(\cos x + \sin x) = n + \frac{1}{2} \]
Thus,
\[ \cos x + \sin x = \frac{n + \frac{1}{2}}{7} \]

### Step 4: Find the Range of \( \cos x + \sin x \)
The maximum value of \( \cos x + \sin x \) occurs when \( x = \frac{\pi}{4} \) and is given by:
\[ \sqrt{2} \quad (\text{since } \cos x + \sin x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right)) \]
The minimum value occurs when \( x = \frac{5\pi}{4} \) and is:
\[ -\sqrt{2} \]
Thus, we have:
\[ -\sqrt{2} \leq \frac{n + \frac{1}{2}}{7} \leq \sqrt{2} \]

### Step 5: Solve for \( n \)
Multiplying through by 7 gives:
\[ -7\sqrt{2} \leq n + \frac{1}{2} \leq 7\sqrt{2} \]
Subtracting \( \frac{1}{2} \) from all parts:
\[ -7\sqrt{2} - \frac{1}{2} \leq n \leq 7\sqrt{2} - \frac{1}{2} \]
Calculating \( 7\sqrt{2} \):
\[ 7\sqrt{2} \approx 7 \times 1.414 \approx 9.898 \]
Thus, we have:
\[ -7\sqrt{2} - \frac{1}{2} \approx -9.898 - 0.5 \approx -10.398 \]
\[ 7\sqrt{2} - \frac{1}{2} \approx 9.898 - 0.5 \approx 9.398 \]
This gives us:
\[ -10.398 \leq n \leq 9.398 \]
The integer values of \( n \) range from \( -10 \) to \( 9 \).

### Step 6: Count the Integer Solutions
The integer values of \( n \) are:
\[ -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 \]
This gives us a total of \( 20 \) values for \( n \).

### Step 7: Calculate the Number of Principal Solutions
Since each value of \( n \) gives us \( 2 \) principal solutions (as \( \sin(x + \frac{\pi}{4}) \) can take values in the range), the total number of principal solutions \( k \) is:
\[ k = 2 \times 20 = 40 \]

### Step 8: Final Calculation
We need to find \( \frac{k}{5} \):
\[ \frac{40}{5} = 8 \]

Thus, the final answer is:
\[ \boxed{8} \]