If `"log"_("cos"x) "tan" x + "log"_("sin"x) "cot" x =0,` then x =

Options

A. `n pi + (pi)/(4), n in Z`
B. `2n pi + (pi)/(4), n in Z`
C. `2n pi -(3pi)/(4), n in Z`
D. none of these

Explanation

To solve the equation \( \log_{\cos x} \tan x + \log_{\sin x} \cot x = 0 \), we can follow these steps:

### Step 1: Rewrite the logarithmic expressions
Using the property of logarithms, we can rewrite \( \log_{\sin x} \cot x \) as \( \log_{\sin x} \frac{1}{\tan x} \):
\[ \log_{\sin x} \cot x = \log_{\sin x} \frac{1}{\tan x} = -\log_{\sin x} \tan x \]
Thus, the equation becomes:
\[ \log_{\cos x} \tan x - \log_{\sin x} \tan x = 0 \]

### Step 2: Combine the logarithmic terms
We can combine the logarithmic terms:
\[ \log_{\cos x} \tan x = \log_{\sin x} \tan x \]
This implies:
\[ \frac{\tan x}{\cos x} = \tan x \]

### Step 3: Simplify the equation
This can be simplified to:
\[ \frac{\tan x}{\cos x} = \tan x \implies \tan x \cdot \cos x = \tan x \]
Assuming \( \tan x \neq 0 \), we can divide both sides by \( \tan x \):
\[ \cos x = 1 \]

### Step 4: Solve for \( x \)
The equation \( \cos x = 1 \) holds true when:
\[ x = 2n\pi \quad (n \in \mathbb{Z}) \]
This means \( x \) can take values like \( 0, 2\pi, 4\pi, \ldots \)

### Step 5: Check for additional solutions
We also need to consider the case when \( \tan x = 0 \):
\[ \tan x = 0 \implies x = n\pi \quad (n \in \mathbb{Z}) \]
This means \( x \) can also take values like \( 0, \pi, 2\pi, \ldots \)

### Conclusion
Thus, the general solutions for \( x \) are:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]