If `log_(cosx) sinx>=2` and `x in [0,3pi]` then `sinx` lies in the interval

Options

A. `[(sqrt5-1)/(2), 1]`
B. `[0(sqrt5-1)/(2)]`
C. `[(1)/(2),1]`
D. none of these

Answer

To solve the inequality \( \log_{\cos x} \sin x \geq 2 \) for \( x \in [0, 3\pi] \) and find the interval in which \( \sin x \) lies, we can follow these steps:

### Step 1: Rewrite the logarithmic inequality
We start with the given inequality:
\[ \log_{\cos x} \sin x \geq 2 \]
Using the change of base formula for logarithms, we can rewrite this as:
\[ \frac{\log \sin x}{\log \cos x} \geq 2 \]

### Step 2: Cross-multiply
Since \( \log \cos x \) is negative for \( x \) in the interval where \( \cos x \) is positive (i.e., \( x \in [0, \frac{\pi}{2}) \cup (2\pi, \frac{5\pi}{2}) \)), we can cross-multiply while flipping the inequality:
\[ \log \sin x \geq 2 \log \cos x \]
This can be rewritten as:
\[ \log \sin x \geq \log (\cos^2 x) \]

### Step 3: Exponentiate both sides
Exponentiating both sides gives us:
\[ \sin x \geq \cos^2 x \]

### Step 4: Substitute \( \cos^2 x \)
We know that \( \cos^2 x = 1 - \sin^2 x \). Therefore, we can rewrite the inequality as:
\[ \sin x \geq 1 - \sin^2 x \]

### Step 5: Rearrange the inequality
Rearranging gives us:
\[ \sin^2 x + \sin x - 1 \geq 0 \]
Let \( a = \sin x \). The inequality becomes:
\[ a^2 + a - 1 \geq 0 \]

### Step 6: Solve the quadratic equation
To find the roots of the quadratic equation \( a^2 + a - 1 = 0 \), we use the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]
Thus, the roots are:
\[ a_1 = \frac{-1 - \sqrt{5}}{2}, \quad a_2 = \frac{-1 + \sqrt{5}}{2} \]

### Step 7: Analyze the intervals
The roots divide the number line into intervals. We need to determine where the quadratic \( a^2 + a - 1 \) is non-negative. The roots are approximately:
- \( a_1 \approx -1.618 \) (not relevant since \( \sin x \) must be between -1 and 1)
- \( a_2 \approx 0.618 \)

The quadratic opens upwards, so it is non-negative outside the roots:
1. \( a \leq \frac{-1 - \sqrt{5}}{2} \) (not relevant)
2. \( a \geq \frac{-1 + \sqrt{5}}{2} \)

### Step 8: Determine the valid range for \( \sin x \)
Since \( \sin x \) must be in the range \([-1, 1]\), we find:
\[ \sin x \in \left[\frac{-1 + \sqrt{5}}{2}, 1\right] \]
Calculating \( \frac{-1 + \sqrt{5}}{2} \):
\[ \frac{-1 + \sqrt{5}}{2} \approx 0.618 \]
Thus, the interval for \( \sin x \) is:
\[ \sin x \in \left[\frac{-1 + \sqrt{5}}{2}, 1\right] \]

### Final Answer:
The interval in which \( \sin x \) lies is:
\[ \left[\frac{-1 + \sqrt{5}}{2}, 1\right] \]