Find the solution set for , ` 4 sin^(2)x -8 sin x +3 le 0` where ` x in [0,2pi]`

Options

A. `[0, pi//6]`
B. `[0, 5pi//6]`
C. `[5pi//6, 2pi]`
D. `[pi//6, 5pi//6]`

Explanation

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

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

### Step 2: Factor the Quadratic Expression
Next, we can factor the quadratic expression. We will look for two numbers that multiply to \( 4 \times 3 = 12 \) and add to \( -8 \). The numbers are \( -6 \) and \( -2 \). Thus, we can rewrite the expression as:
\[ 4 \sin^2 x - 6 \sin x - 2 \sin x + 3 \leq 0 \]
Now, we can factor by grouping:
\[ 2 \sin x (2 \sin x - 3) - 1(2 \sin x - 3) \leq 0 \]
This gives us:
\[ (2 \sin x - 3)(2 \sin x - 1) \leq 0 \]

### Step 3: Identify Critical Points
Now, we need to find the critical points where the expression equals zero:
1. \( 2 \sin x - 3 = 0 \) → \( \sin x = \frac{3}{2} \) (not possible since \( \sin x \) cannot exceed 1)
2. \( 2 \sin x - 1 = 0 \) → \( \sin x = \frac{1}{2} \)

### Step 4: Solve for \( x \)
The solutions for \( \sin x = \frac{1}{2} \) in the interval \( [0, 2\pi] \) are:
\[ x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6} \]

### Step 5: Test Intervals
Next, we will test the intervals determined by the critical points:
1. \( (-\infty, \frac{\pi}{6}) \)
2. \( \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \)
3. \( \left(\frac{5\pi}{6}, 2\pi\right) \)

We can choose test points from each interval:
- For \( x = 0 \) (in \( (-\infty, \frac{\pi}{6}) \)):
\[ (2 \sin 0 - 3)(2 \sin 0 - 1) = (-3)(-1) = 3 > 0 \]
- For \( x = \frac{\pi}{4} \) (in \( \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \)):
\[ (2 \sin \frac{\pi}{4} - 3)(2 \sin \frac{\pi}{4} - 1) = (2 \cdot \frac{\sqrt{2}}{2} - 3)(2 \cdot \frac{\sqrt{2}}{2} - 1) < 0 \]
- For \( x = \frac{3\pi}{2} \) (in \( \left(\frac{5\pi}{6}, 2\pi\right) \)):
\[ (2 \sin \frac{3\pi}{2} - 3)(2 \sin \frac{3\pi}{2} - 1) = (-3)(-7) > 0 \]

### Step 6: Conclusion
The inequality \( (2 \sin x - 3)(2 \sin x - 1) \leq 0 \) holds true in the interval:
\[ \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \]

Thus, the solution set is:
\[ \boxed{\left[\frac{\pi}{6}, \frac{5\pi}{6}\right]} \]