If `2 sin^(2)(x-(pi)/(3))-5sin(x-(pi)/(3))+2lt0`, then belongs to

Options

A. `(((12n-5)pi)/(6),((4n+1)pi)/(2)),n in Z`
B. `(((6n-7)pi)/(6),((2n+1)pi)/(2)),n in Z`
C. `(((4n+1)pi)/(6),n pi),n in Z`
D. `((4n+1)(pi)/(2),(12n+7)(pi)/(6)),n in Z`

Explanation

To solve the inequality \( 2 \sin^2\left(x - \frac{\pi}{3}\right) - 5 \sin\left(x - \frac{\pi}{3}\right) + 2 < 0 \), we will follow these steps:

### Step 1: Substitute \( t \)
Let \( t = \sin\left(x - \frac{\pi}{3}\right) \). Then the inequality becomes:
\[ 2t^2 - 5t + 2 < 0 \]

### Step 2: Factor the quadratic
We need to factor the quadratic expression \( 2t^2 - 5t + 2 \). We can rewrite it as:
\[ 2t^2 - 4t - t + 2 < 0 \]
Grouping the terms:
\[ 2t(t - 2) - 1(t - 2) < 0 \]
Factoring out \( (t - 2) \):
\[ (2t - 1)(t - 2) < 0 \]

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

### Step 4: Determine the intervals
The critical points divide the number line into intervals. We will test the sign of the expression \( (2t - 1)(t - 2) \) in the intervals:
- \( (-\infty, \frac{1}{2}) \)
- \( \left(\frac{1}{2}, 2\right) \)
- \( (2, \infty) \)

### Step 5: Test the intervals
1. For \( t < \frac{1}{2} \) (e.g., \( t = 0 \)):
\[ (2(0) - 1)(0 - 2) = (-1)(-2) = 2 > 0 \]
2. For \( \frac{1}{2} < t < 2 \) (e.g., \( t = 1 \)):
\[ (2(1) - 1)(1 - 2) = (1)(-1) = -1 < 0 \]
3. For \( t > 2 \) (e.g., \( t = 3 \)):
\[ (2(3) - 1)(3 - 2) = (6 - 1)(1) = 5 > 0 \]

### Step 6: Conclusion from intervals
The inequality \( (2t - 1)(t - 2) < 0 \) holds in the interval:
\[ \frac{1}{2} < t < 2 \]
Since \( t = \sin\left(x - \frac{\pi}{3}\right) \), and knowing that the sine function has a maximum value of 1, we restrict \( t \) to:
\[ \frac{1}{2} < \sin\left(x - \frac{\pi}{3}\right) \leq 1 \]

### Step 7: Solve for \( x \)
Now we solve the inequalities:
1. \( \sin\left(x - \frac{\pi}{3}\right) = \frac{1}{2} \)
- This occurs at \( x - \frac{\pi}{3} = \frac{\pi}{6} \) or \( x - \frac{\pi}{3} = \frac{5\pi}{6} \).
- Thus, \( x = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} \) and \( x = \frac{5\pi}{6} + \frac{\pi}{3} = \frac{7\pi}{6} \).

2. The general solutions for \( x \) are:
- \( x = 2n\pi + \frac{\pi}{2} \)
- \( x = 2n\pi + \frac{7\pi}{6} \)

### Final Result
Thus, the final solution for \( x \) is:
\[ x \in \left(2n\pi + \frac{\pi}{2}, 2n\pi + \frac{7\pi}{6}\right) \quad \text{where } n \in \mathbb{Z} \]