If `theta in (0, 2pi)` and `2sin^2theta - 5sintheta + 2 > 0`, then the range of `theta` is

Options

A. `x=2npi + pi/8, n int I`
B. `x=npi + pi/12, n int I`
C. `x=2npi + pi/24, n int l`
D. None of these

Answer

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

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

### Step 2: Factor the quadratic expression
We will factor the quadratic expression \( 2\sin^2\theta - 5\sin\theta + 2 \). To do this, 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 as:
\[ 2\sin^2\theta - 4\sin\theta - \sin\theta + 2 > 0 \]
Now, we can factor by grouping:
\[ 2\sin\theta(\sin\theta - 2) - 1(\sin\theta - 2) > 0 \]
This gives us:
\[ (2\sin\theta - 1)(\sin\theta - 2) > 0 \]

### Step 3: Analyze the factors
We need to find the critical points where each factor is zero:
1. \( 2\sin\theta - 1 = 0 \) implies \( \sin\theta = \frac{1}{2} \)
2. \( \sin\theta - 2 = 0 \) implies \( \sin\theta = 2 \) (not possible since sine values range from -1 to 1)

The only critical point we need to consider is \( \sin\theta = \frac{1}{2} \).

### Step 4: Determine the angles
The angles where \( \sin\theta = \frac{1}{2} \) in the interval \( (0, 2\pi) \) are:
\[ \theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{5\pi}{6} \]

### Step 5: Test intervals
We will test the intervals determined by the critical point \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \):
1. Interval \( (0, \frac{\pi}{6}) \)
2. Interval \( (\frac{\pi}{6}, \frac{5\pi}{6}) \)
3. Interval \( (\frac{5\pi}{6}, 2\pi) \)

- For \( \theta \in (0, \frac{\pi}{6}) \): Choose \( \theta = 0 \) → \( (2(0) - 1)(0 - 2) = (-1)(-2) > 0 \) (True)
- For \( \theta \in (\frac{\pi}{6}, \frac{5\pi}{6}) \): Choose \( \theta = \frac{\pi}{2} \) → \( (2(1) - 1)(1 - 2) = (1)(-1) < 0 \) (False)
- For \( \theta \in (\frac{5\pi}{6}, 2\pi) \): Choose \( \theta = \frac{3\pi}{2} \) → \( (2(-1) - 1)(-1 - 2) = (-3)(-3) > 0 \) (True)

### Step 6: Combine the intervals
The solution to the inequality \( 2\sin^2\theta - 5\sin\theta + 2 > 0 \) is:
\[ \theta \in (0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, 2\pi) \]

### Final Answer
Thus, the range of \( \theta \) is:
\[ \theta \in \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 2\pi\right) \]