The solution of the inequality `log_(1//2) sinx gt log_(1//2) cosx` is

Options

A. `x int (0,pi/2)`
B. `x int (0,pi/2)`
C. `x int (0, pi/4)`
D. `x int [0, pi/4]`

Answer

To solve the inequality \( \log_{(1/2)} \sin x > \log_{(1/2)} \cos x \), we can follow these steps:

### Step 1: Rewrite the inequality
We start by rewriting the given inequality:
\[ \log_{(1/2)} \sin x - \log_{(1/2)} \cos x > 0 \]

### Step 2: Apply the logarithmic property
Using the property of logarithms that states \( \log_a m - \log_a n = \log_a \left(\frac{m}{n}\right) \), we can rewrite the inequality as:
\[ \log_{(1/2)} \left(\frac{\sin x}{\cos x}\right) > 0 \]

### Step 3: Convert the logarithmic inequality
The logarithm \( \log_{(1/2)} y > 0 \) implies that \( y < 1 \) because the base \( \frac{1}{2} \) is less than 1. Therefore, we have:
\[ \frac{\sin x}{\cos x} < 1 \]

### Step 4: Simplify the inequality
The expression \( \frac{\sin x}{\cos x} \) is equal to \( \tan x \). Thus, we can rewrite the inequality as:
\[ \tan x < 1 \]

### Step 5: Solve for \( x \)
The inequality \( \tan x < 1 \) is satisfied in the interval where \( x \) is less than \( \frac{\pi}{4} \) (since \( \tan \frac{\pi}{4} = 1 \)). Therefore, we have:
\[ x < \frac{\pi}{4} \]

### Step 6: Determine the interval
Since \( \tan x \) is defined for \( x \) in the interval \( [0, \frac{\pi}{2}) \), we conclude that the solution to the inequality is:
\[ x \in [0, \frac{\pi}{4}) \]

### Final Answer
The solution of the inequality \( \log_{(1/2)} \sin x > \log_{(1/2)} \cos x \) is:
\[ x \in [0, \frac{\pi}{4}) \]

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