The solution of the equation
`"log"_("cos"x) "sin" x + "log"_("sin"x) "cos" x = 2` is given by

Options

A. `x = n pi + (pi)/(4), n in Z`
B. `x = n pi + (pi)/(2), n in Z`
C. `x = n pi + (pi)/(8), n in Z`
D. `x = 2n pi + (pi)/(6), n in Z`

Answer

To solve the equation

\[ \log_{\cos x} \sin x + \log_{\sin x} \cos x = 2, \]

we will follow these steps:

### Step 1: Apply the Change of Base Formula
Using the property of logarithms, we can rewrite the logarithms in terms of natural logarithms:

\[ \log_{\cos x} \sin x = \frac{\log \sin x}{\log \cos x} \]
and
\[ \log_{\sin x} \cos x = \frac{\log \cos x}{\log \sin x}. \]

Thus, the equation becomes:

\[ \frac{\log \sin x}{\log \cos x} + \frac{\log \cos x}{\log \sin x} = 2. \]

### Step 2: Combine the Terms
Next, we find a common denominator for the left-hand side:

\[ \frac{(\log \sin x)^2 + (\log \cos x)^2}{\log \sin x \cdot \log \cos x} = 2. \]

### Step 3: Cross-Multiply
Cross-multiplying gives us:

\[ (\log \sin x)^2 + (\log \cos x)^2 = 2 \log \sin x \cdot \log \cos x. \]

### Step 4: Rearranging the Equation
Rearranging the equation, we have:

\[ (\log \sin x)^2 - 2 \log \sin x \cdot \log \cos x + (\log \cos x)^2 = 0. \]

### Step 5: Recognize the Perfect Square
This can be recognized as a perfect square:

\[ (\log \sin x - \log \cos x)^2 = 0. \]

### Step 6: Solve the Perfect Square
Taking the square root of both sides gives:

\[ \log \sin x - \log \cos x = 0. \]

This implies:

\[ \log \sin x = \log \cos x. \]

### Step 7: Exponentiate Both Sides
Exponentiating both sides leads to:

\[ \frac{\sin x}{\cos x} = 1, \]

which means:

\[ \tan x = 1. \]

### Step 8: General Solution for Tangent
The general solution for \( \tan x = 1 \) is:

\[ x = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z}. \]

### Final Answer
Thus, the solution of the equation is:

\[ x = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z}. \]

---