The set of all x in the interval `[0, pi]` for which `2 sin^(2)x-3 sin x + 1 ge 0` is

Options

A. `{(pi)/(2)}`
B. `phi`
C. `[0,(pi)/(4)]`
D. `[0,(pi)/(6)]uu[(5pi)/(6),pi]uu{(pi)/(2)}`

Answer

To solve the inequality \(2 \sin^2 x - 3 \sin x + 1 \geq 0\) for \(x\) in the interval \([0, \pi]\), we will follow these steps:

### Step 1: Substitute \(y = \sin x\)
We start by substituting \(y = \sin x\). This transforms our inequality into a quadratic form:
\[ 2y^2 - 3y + 1 \geq 0 \]

### Step 2: Find the roots of the quadratic equation
Next, we need to find the roots of the quadratic equation \(2y^2 - 3y + 1 = 0\) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 2\), \(b = -3\), and \(c = 1\):
\[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
This gives us the roots:
\[ y_1 = \frac{4}{4} = 1 \quad \text{and} \quad y_2 = \frac{2}{4} = \frac{1}{2} \]

### Step 3: Analyze the sign of the quadratic
Now we have the roots \(y = \frac{1}{2}\) and \(y = 1\). We will analyze the sign of the quadratic \(2y^2 - 3y + 1\) in the intervals determined by these roots: \((-\infty, \frac{1}{2})\), \((\frac{1}{2}, 1)\), and \((1, \infty)\).

1. **Interval \((-\infty, \frac{1}{2})\)**: Choose \(y = 0\):
\[ 2(0)^2 - 3(0) + 1 = 1 \quad (\text{positive}) \]

2. **Interval \((\frac{1}{2}, 1)\)**: Choose \(y = \frac{3}{4}\):
\[ 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 1 = 2 \cdot \frac{9}{16} - \frac{9}{4} + 1 = \frac{18}{16} - \frac{36}{16} + \frac{16}{16} = -\frac{2}{16} \quad (\text{negative}) \]

3. **Interval \((1, \infty)\)**: Choose \(y = 2\):
\[ 2(2)^2 - 3(2) + 1 = 8 - 6 + 1 = 3 \quad (\text{positive}) \]

### Step 4: Combine the results
The quadratic \(2y^2 - 3y + 1 \geq 0\) is satisfied in the intervals:
\[ (-\infty, \frac{1}{2}] \cup [1, \infty) \]

### Step 5: Translate back to \(x\)
Since \(y = \sin x\), we need to find the values of \(x\) for which:
1. \(\sin x \leq \frac{1}{2}\)
2. \(\sin x = 1\)

- For \(\sin x \leq \frac{1}{2}\) in the interval \([0, \pi]\):
- \(\sin x = \frac{1}{2}\) occurs at \(x = \frac{\pi}{6}\).
- Thus, \(x\) can take values in the interval \([0, \frac{\pi}{6}]\).

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

### Step 6: Final intervals
Combining these intervals, we find that the solution to the inequality \(2 \sin^2 x - 3 \sin x + 1 \geq 0\) in the interval \([0, \pi]\) is:
\[ x \in [0, \frac{\pi}{6}] \cup [\frac{\pi}{2}, \pi] \]

### Final Answer
The set of all \(x\) in the interval \([0, \pi]\) for which \(2 \sin^2 x - 3 \sin x + 1 \geq 0\) is:
\[ [0, \frac{\pi}{6}] \cup [\frac{\pi}{2}, \pi] \]