If `4 sin^(2) x-8sin x+ 3 le 0,0le x le 2pi` then the solution set for x is

Options

A. `(0,pi)`
B. `(0,(pi)/(4))`
C. `((pi)/(4),pi)`
D. `((pi)/(6),(5pi)/(6))`

Answer

To solve the inequality \(4 \sin^2 x - 8 \sin x + 3 \leq 0\) for \(0 \leq x \leq 2\pi\), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the given inequality:
\[ 4 \sin^2 x - 8 \sin x + 3 \leq 0 \]

### Step 2: Factor the quadratic expression
This is a quadratic inequality in terms of \(\sin x\). We can factor it:
\[ 4 \sin^2 x - 8 \sin x + 3 = (2 \sin x - 3)(2 \sin x - 1) \]
Thus, we rewrite the inequality as:
\[ (2 \sin x - 3)(2 \sin x - 1) \leq 0 \]

### Step 3: Find the critical points
To find the critical points, we set each factor to zero:
1. \(2 \sin x - 3 = 0 \Rightarrow \sin x = \frac{3}{2}\) (not possible since \(\sin x\) cannot exceed 1)
2. \(2 \sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{2}\)

The only valid critical point is \(\sin x = \frac{1}{2}\).

### Step 4: Determine the intervals
Next, we analyze the sign of the product \((2 \sin x - 3)(2 \sin x - 1)\) in the intervals determined by the critical points. The relevant intervals are:
- \((-\infty, \frac{1}{2})\)
- \((\frac{1}{2}, \infty)\)

### Step 5: Test intervals
1. For \(\sin x < \frac{1}{2}\) (e.g., \(\sin x = 0\)):
\((2 \cdot 0 - 3)(2 \cdot 0 - 1) = (-3)(-1) > 0\) (positive)

2. For \(\sin x = \frac{1}{2}\):
\((2 \cdot \frac{1}{2} - 3)(2 \cdot \frac{1}{2} - 1) = (1 - 3)(1 - 1) = (-2)(0) = 0\) (zero)

3. For \(\sin x > \frac{1}{2}\) (e.g., \(\sin x = 1\)):
\((2 \cdot 1 - 3)(2 \cdot 1 - 1) = (2 - 3)(2 - 1) = (-1)(1) < 0\) (negative)

### Step 6: Combine the results
The inequality \((2 \sin x - 3)(2 \sin x - 1) \leq 0\) holds in the interval:
\[ \frac{1}{2} \leq \sin x \leq 1 \]

### Step 7: Find the angles corresponding to \(\sin x = \frac{1}{2}\) and \(\sin x = 1\)
- \(\sin x = \frac{1}{2}\) gives \(x = \frac{\pi}{6}, \frac{5\pi}{6}\)
- \(\sin x = 1\) gives \(x = \frac{\pi}{2}\)

### Step 8: Write the solution set
Thus, the solution set for \(x\) in the interval \(0 \leq x \leq 2\pi\) is:
\[ x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \cup \left[\frac{\pi}{2}, \frac{\pi}{2}\right] \]

### Final Solution
The final solution set for \(x\) is:
\[ \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \]