If the angle between the vectors `vec(a)=hat(i)+(cos x)hat(j)+hat(k)` and
`vec(b) =(sin^(2)x-sin x)hat(i)-(cos x)hat(j)+(3-4sin x)hat(k)`
is obutse and x in `(0,(pi)/(2))`, then the exhaustive set of values of 'x' is equal to-

Options

A. `x in(0, (pi)/(6))`
B. `x in((pi)/(6), (pi)/(2))`
C. `x in((pi)/(6), (pi)/(3))`
D. `x in((pi)/(3), (pi)/(2))`

Answer

To solve the problem, we need to find the values of \( x \) such that the angle between the vectors \( \vec{a} \) and \( \vec{b} \) is obtuse. This means that the dot product of the two vectors must be negative.

### Step-by-Step Solution

1. **Identify the vectors**:
\[ \vec{a} = \hat{i} + \cos x \hat{j} + \hat{k} \]
\[ \vec{b} = (\sin^2 x - \sin x) \hat{i} - \cos x \hat{j} + (3 - 4\sin x) \hat{k} \]

2. **Calculate the dot product \( \vec{a} \cdot \vec{b} \)**:
\[ \vec{a} \cdot \vec{b} = (1)(\sin^2 x - \sin x) + (\cos x)(- \cos x) + (1)(3 - 4\sin x) \]
Simplifying this:
\[ = \sin^2 x - \sin x - \cos^2 x + 3 - 4\sin x \]
Using the identity \( \cos^2 x = 1 - \sin^2 x \):
\[ = \sin^2 x - \sin x - (1 - \sin^2 x) + 3 - 4\sin x \]
\[ = 2\sin^2 x - 5\sin x + 2 \]

3. **Set the dot product less than zero** (since the angle is obtuse):
\[ 2\sin^2 x - 5\sin x + 2 < 0 \]

4. **Let \( t = \sin x \)**, then we have:
\[ 2t^2 - 5t + 2 < 0 \]

5. **Factor the quadratic**:
\[ 2t^2 - 5t + 2 = (2t - 1)(t - 2) \]
Thus, we need to solve:
\[ (2t - 1)(t - 2) < 0 \]

6. **Find the critical points**:
The critical points are \( t = \frac{1}{2} \) and \( t = 2 \). Since \( t = \sin x \) and \( \sin x \) can only take values in the range \( [0, 1] \), we only consider \( t = \frac{1}{2} \).

7. **Determine the intervals**:
The intervals to test are:
- \( t < \frac{1}{2} \)
- \( \frac{1}{2} < t < 2 \) (but since \( t \) cannot be greater than 1, we only consider \( \frac{1}{2} < t < 1 \))
- \( t > 2 \) (not applicable)

Testing the interval \( \left( \frac{1}{2}, 1 \right) \):
- Choose \( t = 0.6 \):
\[ (2(0.6) - 1)(0.6 - 2) = (1.2 - 1)(0.6 - 2) = (0.2)(-1.4) < 0 \]
- Thus, the inequality holds for \( \frac{1}{2} < t < 1 \).

8. **Convert back to \( x \)**:
Since \( t = \sin x \), we have:
\[ \frac{1}{2} < \sin x < 1 \]
This corresponds to:
\[ x \in \left( \frac{\pi}{6}, \frac{\pi}{2} \right) \]

### Final Answer
The exhaustive set of values of \( x \) is:
\[ x \in \left( \frac{\pi}{6}, \frac{\pi}{2} \right) \]