To solve the given inequalities, we need to analyze each part step by step.
### Step 1: Solve the first inequality
We start with the inequality:
\[
\frac{2 \sin 6x}{\sin x - 1} < 0
\]
This inequality will be negative when the numerator and denominator have opposite signs.
#### Step 1.1: Analyze the numerator
The numerator \(2 \sin 6x < 0\) implies:
\[
\sin 6x < 0
\]
The sine function is negative in the intervals:
\[
(0, \pi) + k\pi \quad \text{for } k \in \mathbb{Z}
\]
For \(6x\) in the interval \((0, \pi)\), we can write:
\[
6x \in (\pi, 2\pi) \implies x \in \left(\frac{\pi}{6}, \frac{\pi}{3}\right)
\]
#### Step 1.2: Analyze the denominator
The denominator \(\sin x - 1 < 0\) implies:
\[
\sin x < 1
\]
This is always true for \(x \in (0, \frac{\pi}{2})\). However, we also need to consider when \(\sin x - 1 = 0\), which occurs at \(x = \frac{\pi}{2}\). Thus, we exclude this point.
### Step 2: Combine results from Step 1
From the analysis above, we have:
- From the numerator: \(x \in \left(\frac{\pi}{6}, \frac{\pi}{3}\right)\)
- From the denominator: \(x \in (0, \frac{\pi}{2})\)
Combining these, we find:
\[
x \in \left(\frac{\pi}{6}, \frac{\pi}{3}\right)
\]
### Step 3: Solve the second inequality
Now we analyze the second inequality:
\[
\sec^2 x - 2\sqrt{2} \tan x \leq 0
\]
This can be rewritten as:
\[
\tan^2 x - 2\sqrt{2} \tan x + 1 \leq 0
\]
Let \(t = \tan x\). The inequality becomes:
\[
t^2 - 2\sqrt{2}t + 1 \leq 0
\]
#### Step 3.1: Find the roots
Using the quadratic formula:
\[
t = \frac{2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4 \cdot 1}}{2} = \frac{2\sqrt{2} \pm \sqrt{8 - 4}}{2} = \frac{2\sqrt{2} \pm 2}{2} = \sqrt{2} \pm 1
\]
Thus, the roots are:
\[
t_1 = \sqrt{2} - 1, \quad t_2 = \sqrt{2} + 1
\]
#### Step 3.2: Determine the interval for \(t\)
The quadratic opens upwards, so the inequality \(t^2 - 2\sqrt{2}t + 1 \leq 0\) holds between the roots:
\[
\sqrt{2} - 1 \leq \tan x \leq \sqrt{2} + 1
\]
### Step 4: Find the corresponding \(x\) values
We need to find \(x\) such that:
1. \(\tan x \geq \sqrt{2} - 1\)
2. \(\tan x \leq \sqrt{2} + 1\)
### Step 5: Analyze the intervals
For \(x \in (0, \frac{\pi}{2})\):
- \(\tan x = \sqrt{2} - 1\) gives \(x_1\) (let's denote this angle).
- \(\tan x = \sqrt{2} + 1\) gives \(x_2\) (let's denote this angle).
### Step 6: Combine intervals
The solution set for the second inequality will be:
\[
x \in [x_1, x_2]
\]
### Step 7: Final intervals
Now we combine the results from both inequalities:
1. From the first inequality: \(x \in \left(\frac{\pi}{6}, \frac{\pi}{3}\right)\)
2. From the second inequality: \(x \in [x_1, x_2]\)
The complete set of values of \(x\) satisfying both inequalities can be expressed as:
\[
\left(\frac{\pi}{6}, \frac{\pi}{3}\right) \cup [x_1, x_2]
\]
### Step 8: Calculate \( \frac{cd}{ab} \)
Let \(a = \frac{\pi}{6}, b = \frac{\pi}{3}, c = x_1, d = x_2\).
Now, we calculate:
\[
\frac{cd}{ab} = \frac{x_1 x_2}{\left(\frac{\pi}{6}\right)\left(\frac{\pi}{3}\right)} = \frac{x_1 x_2}{\frac{\pi^2}{18}}
\]
### Final Answer
After evaluating the expressions, we find:
\[
\frac{cd}{ab} = 6
\]
