The set of values of `theta` satisfying the inequatioin `2sin^(2)theta-5sintheta+2 gt 0, ` where `o lt theta lt 2pi`, is

Options

A. `(0,(pi)/(6)) cup ((5pi)/(6),2pi)`
B. `[0,(pi)/(6)] cup [(5pi)/(6),2pi]`
C. `[0,(pi)/(3)] cup [(2pi)/(3),2pi]`
D. None of the above

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
The given inequality is:
\[ 2\sin^2\theta - 5\sin\theta + 2 > 0 \]
Let \( x = \sin\theta \). Then, we can rewrite the inequality as:
\[ 2x^2 - 5x + 2 > 0 \]

### Step 2: Find the Roots of the Quadratic Equation
To find the roots of the quadratic equation \( 2x^2 - 5x + 2 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2 \), \( b = -5 \), and \( c = 2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \]

Now, substituting into the quadratic formula:
\[ x = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4} \]
This gives us two roots:
\[ x_1 = \frac{8}{4} = 2 \quad \text{and} \quad x_2 = \frac{2}{4} = \frac{1}{2} \]

### Step 3: Analyze the Sign of the Quadratic
The roots of the equation \( 2x^2 - 5x + 2 = 0 \) are \( x = 2 \) and \( x = \frac{1}{2} \). Since \( \sin\theta \) can only take values between -1 and 1, we only consider the root \( x = \frac{1}{2} \).

Next, we need to determine the intervals where \( 2x^2 - 5x + 2 > 0 \). The parabola opens upwards (since the coefficient of \( x^2 \) is positive).

The intervals to test are:
1. \( (-\infty, \frac{1}{2}) \)
2. \( (\frac{1}{2}, 2) \)

### Step 4: Test Intervals
- For \( x < \frac{1}{2} \) (e.g., \( x = 0 \)):
\[ 2(0)^2 - 5(0) + 2 = 2 > 0 \]
- For \( \frac{1}{2} < x < 2 \) (e.g., \( x = 1 \)):
\[ 2(1)^2 - 5(1) + 2 = 2 - 5 + 2 = -1 < 0 \]
- For \( x > 2 \) (not applicable since \( \sin\theta \) cannot be greater than 1).

Thus, the inequality \( 2\sin^2\theta - 5\sin\theta + 2 > 0 \) holds for:
\[ \sin\theta < \frac{1}{2} \]

### Step 5: Find Corresponding Angles
The values of \( \theta \) for which \( \sin\theta < \frac{1}{2} \) in the interval \( 0 < \theta < 2\pi \) are:
- From \( 0 \) to \( \frac{\pi}{6} \) (where \( \sin\theta = \frac{1}{2} \))
- From \( \frac{5\pi}{6} \) to \( 2\pi \)

### Final Solution
Thus, the set of values of \( \theta \) satisfying the inequality is:
\[ \theta \in \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 2\pi\right) \]