The solution set of the inequation
`"log"_(1//2) "sin" x gt "log"_(1//2) "cos" x "in" [0, 2pi]`, is

Options

A. `(0, pi//2)`
B. `(-pi//4, pi//4)`
C. `(0, pi//4)`
D. none of these

Answer

To solve the inequation \( \log_{1/2}(\sin x) > \log_{1/2}(\cos x) \) in the interval \( [0, 2\pi] \), we can follow these steps:

### Step 1: Rewrite the Inequation
We start with the given inequation:
\[ \log_{1/2}(\sin x) > \log_{1/2}(\cos x) \]
Using the property of logarithms, we can rewrite this as:
\[ \log_{1/2}(\sin x) - \log_{1/2}(\cos x) > 0 \]
This simplifies to:
\[ \log_{1/2}\left(\frac{\sin x}{\cos x}\right) > 0 \]

### Step 2: Change of Base
Since \( \log_{1/2}(a) > 0 \) implies \( a < 1 \) (because the base \( 1/2 < 1 \)), we can deduce:
\[ \frac{\sin x}{\cos x} < 1 \]
This simplifies to:
\[ \tan x < 1 \]

### Step 3: Determine the Range for \( x \)
The inequality \( \tan x < 1 \) means that \( x \) must be less than \( \frac{\pi}{4} \) (since \( \tan\left(\frac{\pi}{4}\right) = 1 \)). Therefore, we have:
\[ x < \frac{\pi}{4} \]

### Step 4: Consider the Interval
We need to consider the interval \( [0, 2\pi] \). The solution set for \( \tan x < 1 \) in this interval can be found by analyzing the behavior of the tangent function:
- In the interval \( [0, \frac{\pi}{4}) \), \( \tan x < 1 \).
- In the interval \( [\pi, \frac{5\pi}{4}) \), \( \tan x < 1 \) again, since \( \tan x \) is periodic with a period of \( \pi \).

### Step 5: Final Solution Set
Thus, the complete solution set for the inequation \( \log_{1/2}(\sin x) > \log_{1/2}(\cos x) \) in the interval \( [0, 2\pi] \) is:
\[ x \in \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \]