The number of solution of
`sin{sin^(-1)(log_(1//2)x)}+2|cos{sin^(-1)(x/2-3/2)}|=0` is

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To solve the equation

\[ \sin\left(\sin^{-1}\left(\log_{\frac{1}{2}} x\right)\right) + 2\left|\cos\left(\sin^{-1}\left(\frac{x}{2} - \frac{3}{2}\right)\right)\right| = 0, \]

we will break it down step by step.

### Step 1: Understanding the range of \(\log_{\frac{1}{2}} x\)

The logarithm \(\log_{\frac{1}{2}} x\) is defined for \(x > 0\). The range of \(\log_{\frac{1}{2}} x\) is from \(-\infty\) to \(0\) as \(x\) varies from \(1\) to \(\infty\).

To find the specific values for \(x\), we need:

\[ \log_{\frac{1}{2}} x \in [-1, 1]. \]

This translates to:

\[ -1 \leq \log_{\frac{1}{2}} x \leq 0. \]

Converting this to exponential form gives:

\[ \frac{1}{2} \leq x < 1. \]

### Step 2: Analyzing the second term \(2|\cos(\sin^{-1}(\frac{x}{2} - \frac{3}{2}))|\)

The expression \(\sin^{-1}(\frac{x}{2} - \frac{3}{2})\) is defined when \(-1 \leq \frac{x}{2} - \frac{3}{2} \leq 1\).

Solving the inequalities:

1. \(\frac{x}{2} - \frac{3}{2} \geq -1\):
\[ \frac{x}{2} \geq \frac{1}{2} \implies x \geq 1. \]

2. \(\frac{x}{2} - \frac{3}{2} \leq 1\):
\[ \frac{x}{2} \leq \frac{5}{2} \implies x \leq 5. \]

Thus, we have:

\[ 1 \leq x \leq 5. \]

### Step 3: Finding the intersection of the ranges

Now we combine the two ranges we found:

1. From \(\log_{\frac{1}{2}} x\): \(\frac{1}{2} \leq x < 1\).
2. From \(\sin^{-1}(\frac{x}{2} - \frac{3}{2})\): \(1 \leq x \leq 5\).

The intersection of these intervals is:

\[ 1 \leq x < 1 \quad \text{(which is empty)}. \]

### Step 4: Analyzing the equation

Since the intersection of the ranges is empty, there are no values of \(x\) that satisfy both conditions. Therefore, the equation has no solutions.

### Conclusion

The number of solutions to the equation is:

\[ \boxed{0}. \]