If `log_(sinx)(cos x) = (1)/(2)`, where `x in (0, (pi)/(2))`, then the value of `sin x` is equal to-

Options

A. `(1)/(sqrt(3))`
B. `(sqrt(5) - 1)/(2)`
C. `(2)/(sqrt(3))`
D. `(sqrt(5) + 1)/(2)`

Explanation

To solve the equation \( \log_{\sin x}(\cos x) = \frac{1}{2} \) where \( x \in (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Rewrite the logarithmic equation
Starting with the equation:
\[ \log_{\sin x}(\cos x) = \frac{1}{2} \]
We can rewrite this using the change of base formula:
\[ \frac{\log(\cos x)}{\log(\sin x)} = \frac{1}{2} \]

### Step 2: Cross-multiply
Cross-multiplying gives:
\[ 2 \log(\cos x) = \log(\sin x) \]

### Step 3: Apply the power rule of logarithms
Using the property of logarithms \( n \log(m) = \log(m^n) \), we can rewrite the equation:
\[ \log(\cos x) = \log((\sin x)^{1/2}) \]

### Step 4: Remove the logarithm
Since the logarithm is one-to-one, we can equate the arguments:
\[ \cos x = (\sin x)^{1/2} \]

### Step 5: Square both sides
Squaring both sides to eliminate the square root gives:
\[ \cos^2 x = \sin x \]

### Step 6: Use the Pythagorean identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can substitute \( \cos^2 x \):
\[ 1 - \sin^2 x = \sin x \]

### Step 7: Rearrange into a standard quadratic form
Rearranging gives us:
\[ \sin^2 x + \sin x - 1 = 0 \]

### Step 8: Apply the quadratic formula
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -1 \):
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ = \frac{-1 \pm \sqrt{1 + 4}}{2} \]
\[ = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 9: Determine the valid solution
Since \( x \in (0, \frac{\pi}{2}) \), \( \sin x \) must be positive. Therefore, we take the positive root:
\[ \sin x = \frac{-1 + \sqrt{5}}{2} \]

### Final Answer
Thus, the value of \( \sin x \) is:
\[ \sin x = \frac{\sqrt{5} - 1}{2} \]