The solution of the equation `log _( cosx ^(2)) (3-2x) lt log _( cos x ^(2)) (2x -1) ` is:

Options

A. `(1//2,1)`
B. `(-oo,1)`
C. `(1//2, 3)`
D. `(1,oo -sqrt(2pi pi,) n in N `

Answer

To solve the inequality \( \log_{\cos^2 x}(3 - 2x) < \log_{\cos^2 x}(2x - 1) \), we will follow these steps:

### Step 1: Understand the properties of logarithms
We know that if \( \log_b(a) < \log_b(c) \), then \( a < c \) if \( b > 1 \) and \( a > c \) if \( 0 < b < 1 \). Here, \( b = \cos^2 x \).

### Step 2: Determine the base of the logarithm
Since \( \cos^2 x \) is always between 0 and 1 for \( x \) in the interval where cosine is defined, we will reverse the inequality:
\[ 3 - 2x > 2x - 1 \]

### Step 3: Solve the inequality
Now, we will solve the inequality:
\[ 3 - 2x > 2x - 1 \]
Rearranging gives:
\[ 3 + 1 > 2x + 2x \]
\[ 4 > 4x \]
Dividing both sides by 4:
\[ 1 > x \quad \text{or} \quad x < 1 \]

### Step 4: Determine the domain of the logarithmic function
Next, we need to ensure that the arguments of the logarithms are positive:
1. \( 3 - 2x > 0 \)
\[ 3 > 2x \quad \Rightarrow \quad x < \frac{3}{2} \]
2. \( 2x - 1 > 0 \)
\[ 2x > 1 \quad \Rightarrow \quad x > \frac{1}{2} \]

### Step 5: Combine the inequalities
Now we have:
- From the logarithmic inequality: \( x < 1 \)
- From the domain conditions: \( \frac{1}{2} < x < \frac{3}{2} \)

### Step 6: Final solution
Combining these inequalities, we find:
\[ \frac{1}{2} < x < 1 \]

Thus, the solution to the inequality \( \log_{\cos^2 x}(3 - 2x) < \log_{\cos^2 x}(2x - 1) \) is:
\[ \boxed{\left( \frac{1}{2}, 1 \right)} \]