The number of ordered pairs `(alpha , beta )` , where ` alpha , beta in [0,2pi]` satisfying ` log_(2 sec x )(beta^(2)-6beta+10)=log_(3)| cos alpha |` is

Answer

To solve the problem, we need to find the number of ordered pairs \((\alpha, \beta)\) such that:

\[ \log_{2}(\sec x)(\beta^{2} - 6\beta + 10) = \log_{3}|\cos \alpha| \]

where \(\alpha, \beta \in [0, 2\pi]\).

### Step 1: Simplifying the logarithmic equation

From the equation, we can equate the arguments of the logarithms:

\[ 2^{\log_{2}(\sec x)} = 3^{\log_{3}|\cos \alpha|} \]

This simplifies to:

\[ \sec x(\beta^{2} - 6\beta + 10) = |\cos \alpha| \]

### Step 2: Analyzing the secant function

The secant function, \(\sec x\), is defined as \( \sec x = \frac{1}{\cos x} \). Thus, we can rewrite the equation as:

\[ \frac{1}{\cos x}(\beta^{2} - 6\beta + 10) = |\cos \alpha| \]

### Step 3: Setting bounds for \(|\cos \alpha|\)

Since \(|\cos \alpha|\) can take values in the range \([0, 1]\), we have:

\[ 0 < \frac{1}{\cos x}(\beta^{2} - 6\beta + 10) \leq 1 \]

This implies:

\[ 0 < \beta^{2} - 6\beta + 10 \leq \cos x \]

### Step 4: Finding the discriminant of the quadratic

Next, we need to analyze the quadratic \( \beta^{2} - 6\beta + 10 \). First, we find its discriminant:

\[ D = b^{2} - 4ac = 6^{2} - 4 \cdot 1 \cdot 10 = 36 - 40 = -4 \]

Since the discriminant is negative, the quadratic does not intersect the x-axis and is always positive. Therefore, \( \beta^{2} - 6\beta + 10 > 0 \) for all \(\beta\).

### Step 5: Setting the upper bound

Next, we need to satisfy the condition:

\[ \beta^{2} - 6\beta + 10 \leq 1 \]

This leads to:

\[ \beta^{2} - 6\beta + 9 \leq 0 \]

Factoring gives:

\[ (\beta - 3)^{2} \leq 0 \]

The only solution to this inequality is:

\[ \beta - 3 = 0 \implies \beta = 3 \]

### Step 6: Finding values of \(\alpha\)

Now substituting \(\beta = 3\) back into the equation:

\[ \beta^{2} - 6\beta + 10 = 3^{2} - 6 \cdot 3 + 10 = 9 - 18 + 10 = 1 \]

Thus, we have:

\[ |\cos \alpha| = 1 \]

The values of \(\alpha\) that satisfy this condition are:

\[ \alpha = 0, \pi, 2\pi \]

### Step 7: Counting the ordered pairs

We have found that \(\beta = 3\) and the possible values of \(\alpha\) are \(0\), \(\pi\), and \(2\pi\). Therefore, the ordered pairs \((\alpha, \beta)\) are:

1. \((0, 3)\)
2. \((\pi, 3)\)
3. \((2\pi, 3)\)

Thus, the total number of ordered pairs \((\alpha, \beta)\) is:

\[ \boxed{3} \]