If `log_(cosx)sin x ge 2 and 0le xle 3pi`, then the value of `sinx` lies in the interval `(0, (sqrta-b)/(2)]` then value of a-b is

Options

A. 3
B. 4
C. 5
D. None of these

Answer

To solve the problem, we need to analyze the inequality given by the logarithmic condition and find the appropriate values for \( a \) and \( b \).

### Step-by-Step Solution:

1. **Understand the Inequality**: We start with the inequality:
\[ \log_{\cos x} \sin x \geq 2 \]
This can be rewritten using the property of logarithms:
\[ \sin x \geq \cos^2 x \]

2. **Express \(\cos^2 x\)**: We know from the Pythagorean identity that:
\[ \cos^2 x = 1 - \sin^2 x \]
Therefore, we can substitute this into our inequality:
\[ \sin x \geq 1 - \sin^2 x \]

3. **Rearranging the Inequality**: Rearranging gives us:
\[ \sin^2 x + \sin x - 1 \geq 0 \]
This is a quadratic inequality in terms of \(\sin x\).

4. **Finding the Roots**: We can find the roots of the quadratic equation:
\[ t^2 + t - 1 = 0 \]
where \( t = \sin x \). Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]
Thus, the roots are:
\[ t_1 = \frac{-1 - \sqrt{5}}{2}, \quad t_2 = \frac{-1 + \sqrt{5}}{2} \]

5. **Analyzing the Quadratic**: The quadratic opens upwards (since the coefficient of \( t^2 \) is positive). Therefore, the inequality \( t^2 + t - 1 \geq 0 \) holds outside the interval defined by the roots:
\[ t \leq \frac{-1 - \sqrt{5}}{2} \quad \text{or} \quad t \geq \frac{-1 + \sqrt{5}}{2} \]

6. **Considering the Range of \(\sin x\)**: Since \(\sin x\) must be in the range \([0, 1]\), we only consider:
\[ \sin x \geq \frac{-1 + \sqrt{5}}{2} \]

7. **Finding the Upper Bound**: We also know that \(\sin x\) must be less than or equal to 1. Therefore, we need to find the interval for \(\sin x\):
\[ 0 < \sin x \leq \frac{-1 + \sqrt{5}}{2} \]

8. **Identifying the Values of \( a \) and \( b \)**: The problem states that \(\sin x\) lies in the interval \( (0, \frac{\sqrt{a} - b}{2}] \). We can compare:
\[ \frac{\sqrt{a} - b}{2} = \frac{-1 + \sqrt{5}}{2} \]
Multiplying through by 2 gives:
\[ \sqrt{a} - b = -1 + \sqrt{5} \]

9. **Solving for \( a \) and \( b \)**: We can set:
\[ \sqrt{a} = \sqrt{5}, \quad b = 1 \]
Thus:
\[ a = 5, \quad b = 1 \]

10. **Finding \( a - b \)**: Finally, we compute:
\[ a - b = 5 - 1 = 4 \]

### Final Answer:
The value of \( a - b \) is \( \boxed{4} \).