Maximum value of the expression
`log_(3)(9-2 cos^(2)theta-4 sec^(2) theta)` is equal to

Answer

To find the maximum value of the expression \( \log_{3}(9 - 2 \cos^{2} \theta - 4 \sec^{2} \theta) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ z = \log_{3}(9 - 2 \cos^{2} \theta - 4 \sec^{2} \theta) \]
We know that \( \sec^{2} \theta = \frac{1}{\cos^{2} \theta} \), so we can rewrite the expression as:
\[ z = \log_{3}(9 - 2 \cos^{2} \theta - \frac{4}{\cos^{2} \theta}) \]

### Step 2: Define a new variable
Let \( x = \cos^{2} \theta \). The expression becomes:
\[ z = \log_{3}(9 - 2x - \frac{4}{x}) \]
where \( 0 < x \leq 1 \).

### Step 3: Find the minimum of the expression inside the logarithm
To maximize \( z \), we need to minimize the expression \( 2x + \frac{4}{x} \). We can denote:
\[ f(x) = 2x + \frac{4}{x} \]

### Step 4: Differentiate and find critical points
We differentiate \( f(x) \):
\[ f'(x) = 2 - \frac{4}{x^2} \]
Setting \( f'(x) = 0 \):
\[ 2 - \frac{4}{x^2} = 0 \implies 2 = \frac{4}{x^2} \implies x^2 = 2 \implies x = \sqrt{2} \]
However, since \( \sqrt{2} > 1 \), we check the endpoints of the interval \( (0, 1] \).

### Step 5: Evaluate \( f(x) \) at the endpoints
1. As \( x \to 0 \), \( f(x) \to \infty \).
2. At \( x = 1 \):
\[ f(1) = 2(1) + \frac{4}{1} = 6 \]

### Step 6: Find the minimum value
Thus, the minimum value of \( 2x + \frac{4}{x} \) occurs at \( x = 1 \) and is equal to 6.

### Step 7: Substitute back to find the maximum value of \( z \)
Now substituting back into our logarithmic expression:
\[ 9 - (2x + \frac{4}{x}) = 9 - 6 = 3 \]
Thus, we have:
\[ z = \log_{3}(3) = 1 \]

### Conclusion
The maximum value of the expression \( \log_{3}(9 - 2 \cos^{2} \theta - 4 \sec^{2} \theta) \) is:
\[ \boxed{1} \]