Set of values of x lying in `[0, 2pi]` satisfying the inequality `|sin x | gt 2 sin^(2)` x contains

Options

A. `(0, pi/6) uu (pi, (7pi)/(6))`
B. `(0, (7pi)/(6))`
C. `pi/6`
D. None of these

Answer

To solve the inequality \( |\sin x| > 2 \sin^2 x \) for \( x \) in the interval \( [0, 2\pi] \), we will follow these steps:

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

### Step 2: Case 1: \( \sin x \geq 0 \)
In this case, \( |\sin x| = \sin x \). Thus, the inequality becomes:
\[ \sin x > 2 \sin^2 x \]
Rearranging gives:
\[ \sin x - 2 \sin^2 x > 0 \]
Factoring out \( \sin x \):
\[ \sin x (1 - 2 \sin x) > 0 \]
This product is positive when both factors are positive or both factors are negative.

### Step 3: Find Critical Points for Case 1
The critical points occur when:
1. \( \sin x = 0 \) which gives \( x = 0, \pi, 2\pi \)
2. \( 1 - 2 \sin x = 0 \) which gives \( \sin x = \frac{1}{2} \) or \( x = \frac{\pi}{6}, \frac{5\pi}{6} \)

### Step 4: Test Intervals for Case 1
The critical points divide the interval \( [0, \pi] \) into the following intervals:
- \( (0, \frac{\pi}{6}) \)
- \( (\frac{\pi}{6}, \frac{5\pi}{6}) \)
- \( (\frac{5\pi}{6}, \pi) \)

Testing these intervals:
1. For \( x \in (0, \frac{\pi}{6}) \): Choose \( x = \frac{\pi}{12} \)
- \( \sin(\frac{\pi}{12} ) > 2 \sin^2(\frac{\pi}{12}) \) → True
2. For \( x \in (\frac{\pi}{6}, \frac{5\pi}{6}) \): Choose \( x = \frac{\pi}{2} \)
- \( \sin(\frac{\pi}{2}) > 2 \sin^2(\frac{\pi}{2}) \) → False
3. For \( x \in (\frac{5\pi}{6}, \pi) \): Choose \( x = \frac{3\pi}{4} \)
- \( \sin(\frac{3\pi}{4}) > 2 \sin^2(\frac{3\pi}{4}) \) → True

Thus, for Case 1, the solution is:
\[ x \in [0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \pi) \]

### Step 5: Case 2: \( \sin x < 0 \)
In this case, \( |\sin x| = -\sin x \). Thus, the inequality becomes:
\[ -\sin x > 2 \sin^2 x \]
Rearranging gives:
\[ 2 \sin^2 x + \sin x > 0 \]
Factoring out \( \sin x \):
\[ \sin x (2 \sin x + 1) > 0 \]

### Step 6: Find Critical Points for Case 2
The critical points occur when:
1. \( \sin x = 0 \) which gives \( x = \pi, 2\pi \)
2. \( 2 \sin x + 1 = 0 \) which gives \( \sin x = -\frac{1}{2} \) or \( x = \frac{7\pi}{6}, \frac{11\pi}{6} \)

### Step 7: Test Intervals for Case 2
The critical points divide the interval \( [\pi, 2\pi] \) into:
- \( (\pi, \frac{7\pi}{6}) \)
- \( (\frac{7\pi}{6}, \frac{11\pi}{6}) \)
- \( (\frac{11\pi}{6}, 2\pi) \)

Testing these intervals:
1. For \( x \in (\pi, \frac{7\pi}{6}) \): Choose \( x = \frac{5\pi}{6} \)
- \( \sin(\frac{5\pi}{6}) < 0 \) → True
2. For \( x \in (\frac{7\pi}{6}, \frac{11\pi}{6}) \): Choose \( x = \frac{3\pi}{2} \)
- \( \sin(\frac{3\pi}{2}) < 0 \) → False
3. For \( x \in (\frac{11\pi}{6}, 2\pi) \): Choose \( x = \frac{13\pi}{6} \)
- \( \sin(\frac{13\pi}{6}) < 0 \) → True

Thus, for Case 2, the solution is:
\[ x \in [\pi, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi) \]

### Step 8: Combine Solutions
Combining the solutions from both cases, we have:
\[ x \in [0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \pi) \cup [\pi, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi) \]

### Final Answer
The set of values of \( x \) satisfying the inequality \( |\sin x| > 2 \sin^2 x \) in the interval \( [0, 2\pi] \) is:
\[ x \in [0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \frac{7\pi}{6}) \cup (\frac{11\pi}{6}, 2\pi) \]