If `0 < theta < 2pi`, then the intervals of values of ? for which `2sin^2theta-5sintheta+2>0`, is

Options

A. `(0, (pi)/(6))uu((5pi)/(6),2pi)`
B. `((pi)/(8),(5pi)/(6))`
C. `(0,(pi)/(8))uu((pi)/(6),(5pi)/(6))`
D. `((41pi)/(48),pi)`

Answer

To solve the inequality \( 2\sin^2\theta - 5\sin\theta + 2 > 0 \) for \( 0 < \theta < 2\pi \), we can follow these steps:

### Step 1: Rewrite the inequality
We start with the inequality:
\[ 2\sin^2\theta - 5\sin\theta + 2 > 0 \]

### Step 2: Let \( x = \sin\theta \)
We can substitute \( x \) for \( \sin\theta \):
\[ 2x^2 - 5x + 2 > 0 \]

### Step 3: Factor the quadratic
Next, we will factor the quadratic expression \( 2x^2 - 5x + 2 \). We look for two numbers that multiply to \( 2 \times 2 = 4 \) and add to \( -5 \). The numbers are \( -4 \) and \( -1 \). Thus, we can rewrite the expression:
\[ 2x^2 - 4x - x + 2 > 0 \]
Grouping the terms gives us:
\[ 2x(x - 2) - 1(x - 2) > 0 \]
Factoring out \( (x - 2) \):
\[ (2x - 1)(x - 2) > 0 \]

### Step 4: Find the critical points
Now we find the critical points by setting each factor to zero:
1. \( 2x - 1 = 0 \) → \( x = \frac{1}{2} \)
2. \( x - 2 = 0 \) → \( x = 2 \)

### Step 5: Determine the intervals
The critical points divide the number line into intervals. We will test the sign of the product \( (2x - 1)(x - 2) \) in each interval:
1. \( (-\infty, \frac{1}{2}) \)
2. \( (\frac{1}{2}, 2) \)
3. \( (2, \infty) \)

### Step 6: Test the intervals
- For \( x < \frac{1}{2} \) (e.g., \( x = 0 \)):
\[ (2(0) - 1)(0 - 2) = (-1)(-2) = 2 > 0 \]
- For \( \frac{1}{2} < x < 2 \) (e.g., \( x = 1 \)):
\[ (2(1) - 1)(1 - 2) = (1)(-1) = -1 < 0 \]
- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ (2(3) - 1)(3 - 2) = (5)(1) = 5 > 0 \]

### Step 7: Identify the solution intervals
The inequality \( (2x - 1)(x - 2) > 0 \) holds in the intervals:
1. \( (-\infty, \frac{1}{2}) \)
2. \( (2, \infty) \)

### Step 8: Translate back to \( \theta \)
Since \( x = \sin\theta \), we need to find the values of \( \theta \) for which:
1. \( \sin\theta < \frac{1}{2} \)
2. \( \sin\theta > 2 \) (not possible since \( \sin\theta \) is always between -1 and 1)

The inequality \( \sin\theta < \frac{1}{2} \) occurs in the intervals:
- \( 0 < \theta < \frac{\pi}{6} \)
- \( \frac{5\pi}{6} < \theta < 2\pi \)

### Final Answer
Thus, the intervals of values of \( \theta \) for which \( 2\sin^2\theta - 5\sin\theta + 2 > 0 \) are:
\[ \theta \in \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 2\pi\right) \]