The number of real solution of the equation
`sin^(-1) (sum_(i=1)^(oo) x^(i +1) -x sum_(i=1)^(oo) ((x)/(2))^(i))`
`= (pi)/(2) - cos^(-1) (sum_(i=1)^(oo) (-(x)/(2))^(i) - sum_(i=1)^(oo) (-x)^(i))` lying in the interval `(-(1)/(2), (1)/(2))` is ______.
(Here, the inverse trigonometric function `sin^(-1) x and cos^(-1) x` assume values in `[-(pi)/(2), (pi)/(2)] and [0, pi]` respectively)

Explanation

To solve the equation

\[ \sin^{-1} \left( \sum_{i=1}^{\infty} x^{i+1} - x \sum_{i=1}^{\infty} \left( \frac{x}{2} \right)^{i} \right) = \frac{\pi}{2} - \cos^{-1} \left( \sum_{i=1}^{\infty} \left( -\frac{x}{2} \right)^{i} - \sum_{i=1}^{\infty} (-x)^{i} \right) \]

we will first simplify the series involved.

### Step 1: Simplifying the series

The series can be recognized as geometric series.

1. **For the first series**:
\[ \sum_{i=1}^{\infty} x^{i+1} = x^2 \sum_{i=0}^{\infty} x^i = x^2 \cdot \frac{1}{1-x} \quad \text{(for } |x| < 1\text{)} \]

2. **For the second series**:
\[ \sum_{i=1}^{\infty} \left( \frac{x}{2} \right)^{i} = \frac{\frac{x}{2}}{1 - \frac{x}{2}} = \frac{x/2}{1 - x/2} = \frac{x}{2 - x} \quad \text{(for } |x| < 2\text{)} \]

3. **For the third series**:
\[ \sum_{i=1}^{\infty} \left( -\frac{x}{2} \right)^{i} = \frac{-\frac{x}{2}}{1 + \frac{x}{2}} = \frac{-x/2}{1 + x/2} = \frac{-x}{2 + x} \quad \text{(for } |x| < 2\text{)} \]

4. **For the fourth series**:
\[ \sum_{i=1}^{\infty} (-x)^{i} = \frac{-x}{1 + x} \quad \text{(for } |x| < 1\text{)} \]

### Step 2: Substitute back into the equation

Now substituting these into the original equation gives:

\[ \sin^{-1} \left( \frac{x^2}{1-x} - x \cdot \frac{x}{2-x} \right) = \frac{\pi}{2} - \cos^{-1} \left( \frac{-x}{2+x} - \frac{-x}{1+x} \right) \]

### Step 3: Simplifying the left-hand side

The left-hand side simplifies to:

\[ \frac{x^2}{1-x} - \frac{x^2}{2-x} = \frac{x^2(2-x) - x^2(1-x)}{(1-x)(2-x)} = \frac{x^2(2-x - 1 + x)}{(1-x)(2-x)} = \frac{x^2}{(1-x)(2-x)} \]

### Step 4: Simplifying the right-hand side

The right-hand side simplifies to:

\[ \frac{-x(1+x) + x(2+x)}{(2+x)(1+x)} = \frac{-x + 2x + x^2}{(2+x)(1+x)} = \frac{x + x^2}{(2+x)(1+x)} \]

### Step 5: Setting the equations equal

Now we have:

\[ \sin^{-1} \left( \frac{x^2}{(1-x)(2-x)} \right) = \frac{\pi}{2} - \cos^{-1} \left( \frac{x + x^2}{(2+x)(1+x)} \right) \]

Using the identity \(\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}\), we can equate the arguments:

\[ \frac{x^2}{(1-x)(2-x)} = \frac{x + x^2}{(2+x)(1+x)} \]

### Step 6: Cross-multiplying and simplifying

Cross-multiplying gives:

\[ x^2(2+x)(1+x) = (x + x^2)(1-x)(2-x) \]

Expanding both sides and simplifying will lead to a polynomial equation.

### Step 7: Finding the roots

After simplifying, we will find a cubic polynomial. We can check the number of real roots using the derivative and the Intermediate Value Theorem.

### Step 8: Conclusion

After analyzing the cubic polynomial, we can conclude that there are 2 real solutions in the interval \((-1/2, 1/2)\).