The set of all `x` in the interval `[0,pi]` for which `2sin^2x-3sinx+1geq0` is______

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: Rewrite the Inequality
We start with the inequality:
\[ 2\sin^2 x - 3\sin x + 1 \geq 0 \]
Let \(y = \sin x\). The inequality becomes:
\[ 2y^2 - 3y + 1 \geq 0 \]

### Step 2: Factor the Quadratic Expression
Next, we factor the quadratic expression:
\[ 2y^2 - 3y + 1 = (2y - 1)(y - 1) \]
Thus, we rewrite the inequality as:
\[ (2y - 1)(y - 1) \geq 0 \]

### Step 3: Find the Roots
To find the critical points, we set each factor to zero:
1. \(2y - 1 = 0 \Rightarrow y = \frac{1}{2}\)
2. \(y - 1 = 0 \Rightarrow y = 1\)

### Step 4: Analyze the Sign of the Expression
We need to analyze the sign of the product \((2y - 1)(y - 1)\) in the intervals determined by the roots:
- The critical points divide the number line into intervals: \((- \infty, \frac{1}{2})\), \((\frac{1}{2}, 1)\), and \((1, \infty)\).

We test a point in each interval:
1. For \(y < \frac{1}{2}\) (e.g., \(y = 0\)):
\((2(0) - 1)(0 - 1) = (-1)(-1) = 1\) (positive)
2. For \(\frac{1}{2} < y < 1\) (e.g., \(y = 0.75\)):
\((2(0.75) - 1)(0.75 - 1) = (0.5)(-0.25) = -0.125\) (negative)
3. For \(y > 1\) (e.g., \(y = 2\)):
\((2(2) - 1)(2 - 1) = (3)(1) = 3\) (positive)

### Step 5: Combine the Results
The inequality \((2y - 1)(y - 1) \geq 0\) holds in the intervals:
- \(y \leq \frac{1}{2}\) or \(y \geq 1\)

### Step 6: Translate Back to \(x\)
Now we translate back to \(x\):
1. \(y \leq \frac{1}{2}\) implies \(\sin x \leq \frac{1}{2}\). The solutions in the interval \([0, \pi]\) are:
- \(x \in [0, \frac{\pi}{6}]\) (where \(\sin x = \frac{1}{2}\))
2. \(y \geq 1\) implies \(\sin x = 1\). The solution is:
- \(x = \frac{\pi}{2}\)

### Step 7: Combine All Solutions
Thus, the complete solution set for \(x\) in the interval \([0, \pi]\) is:
\[ x \in [0, \frac{\pi}{6}] \cup \{\frac{\pi}{2}\} \]

### 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:
\[ [0, \frac{\pi}{6}] \cup \{\frac{\pi}{2}\} \]