The complete set of values of ` x, x in (-(pi)/(2), pi)` satisfying the inequality `cos 2x gt |sin x| ` is :

Options

A. ` (- (pi)/(6), (pi)/(6))`
B. `(-(pi)/(2), (pi)/(6)) cup ((pi)/(6), (5 pi)/(6)) `
C. `(-(pi)/(2), -(pi)/(6)) cup ((5 pi)/(6), pi ) `
D. `(-(pi)/(6),(pi)/(6)) cup ((5 pi)/(6), pi ) `

Answer

To solve the inequality \( \cos 2x > |\sin x| \) for \( x \) in the interval \( \left(-\frac{\pi}{2}, \pi\right) \), we will break the problem down into manageable steps.

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ \cos 2x > |\sin x| \]
This can be split into two cases based on the definition of the absolute value.

### Step 2: Case 1: \( \cos 2x > \sin x \)
For the first case, we rewrite the inequality:
\[ \cos 2x - \sin x > 0 \]
Using the double angle formula for cosine, we have:
\[ \cos 2x = 1 - 2\sin^2 x \]
Substituting this into the inequality gives:
\[ 1 - 2\sin^2 x - \sin x > 0 \]
Rearranging this, we get:
\[ -2\sin^2 x - \sin x + 1 > 0 \]
Multiplying through by -1 (which reverses the inequality):
\[ 2\sin^2 x + \sin x - 1 < 0 \]

### Step 3: Factor the Quadratic
Next, we factor the quadratic:
\[ 2\sin^2 x + \sin x - 1 = 0 \]
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \]
This gives us:
\[ \sin x = \frac{1}{2} \quad \text{and} \quad \sin x = -1 \]

### Step 4: Find the Values of \( x \)
1. For \( \sin x = \frac{1}{2} \):
- \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \)

2. For \( \sin x = -1 \):
- \( x = -\frac{\pi}{2} \) (not included in the interval)

### Step 5: Analyze the Intervals
We need to determine where \( 2\sin^2 x + \sin x - 1 < 0 \) between the critical points \( -\frac{\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}, \pi \).

### Step 6: Test Intervals
1. Test interval \( \left(-\frac{\pi}{2}, \frac{\pi}{6}\right) \)
2. Test interval \( \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \)
3. Test interval \( \left(\frac{5\pi}{6}, \pi\right) \)

After testing these intervals, we find that the solution for this case is:
\[ x \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \pi\right) \]

### Step 7: Case 2: \( \cos 2x > -\sin x \)
For the second case, we rewrite the inequality:
\[ \cos 2x + \sin x > 0 \]
Following similar steps as in Case 1, we arrive at:
\[ 2\sin^2 x - \sin x - 1 < 0 \]
Factoring gives:
\[ (2\sin x + 1)(\sin x - 1) < 0 \]

### Step 8: Find the Values of \( x \)
1. For \( \sin x = -\frac{1}{2} \):
- \( x = -\frac{\pi}{6} \) and \( x = \frac{7\pi}{6} \) (not included in the interval)

2. For \( \sin x = 1 \):
- \( x = \frac{\pi}{2} \)

### Step 9: Analyze the Intervals
We need to determine where \( (2\sin x + 1)(\sin x - 1) < 0 \) between the critical points \( -\frac{\pi}{2}, -\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \pi \).

### Final Solution
Combining the results from both cases, we find the complete set of values of \( x \):
\[ x \in \left(-\frac{\pi}{6}, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \pi\right) \]