The given equation is: $\sin ^{ - 1}\left( {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} - x\sum\limits_{i = 1}^\infty {} {{\left( {{x \over 2}} \right)}^i}} \right) = {\pi \over 2} - {\cos ^{ - 1}}\left( {\sum\limits_{i = 1}^\infty {} {{\left( {{{ - x} \over 2}} \right)}^i} - \sum\limits_{i = 1}^\infty {} {{\left( { - x} \right)}^i}} \right)$

1. Simplify the infinite geometric series. We need to evaluate the four infinite geometric series. The sum of an infinite geometric series with first term $a$ and common ratio $r$ is $S = \frac{a}{1-r}$, provided $|r| < 1$.

   • $S_1 = \sum\limits_{i = 1}^\infty {x^{i + 1}} = x^2 + x^3 + x^4 + \dots = \frac{x^2}{1-x}$ for $|x| < 1$.
   • $S_2 = \sum\limits_{i = 1}^\infty {\left( \frac{x}{2} \right)^i} = \frac{x}{2} + \left( \frac{x}{2} \right)^2 + \dots = \frac{x/2}{1 - x/2} = \frac{x}{2-x}$ for $|x/2| < 1 \implies |x| < 2$.
   • $S_3 = \sum\limits_{i = 1}^\infty {\left( \frac{-x}{2} \right)^i} = \frac{-x}{2} + \left( \frac{-x}{2} \right)^2 + \dots = \frac{-x/2}{1 - (-x/2)} = \frac{-x}{2+x}$ for $|-x/2| < 1 \implies |x| < 2$.
   • $S_4 = \sum\limits_{i = 1}^\infty {(-x)^i} = -x + (-x)^2 + \dots = \frac{-x}{1 - (-x)} = \frac{-x}{1+x}$ for $|-x| < 1 \implies |x| < 1$.

   The problem asks for solutions in the interval $\left( -\frac{1}{2}, \frac{1}{2} \right)$, which is within the convergence region $|x|<1$ for all series.

2. Rewrite the equation with the sums. Let the argument of $\sin^{-1}$ be $A$ and the argument of $\cos^{-1}$ be $B$. $A = S_1 - x S_2 = \frac{x^2}{1-x} - x\left(\frac{x}{2-x}\right) = \frac{x^2(2-x) - x^2(1-x)}{(1-x)(2-x)} = \frac{2x^2 - x^3 - x^2 + x^3}{(1-x)(2-x)} = \frac{x^2}{(1-x)(2-x)}$. $B = S_3 - S_4 = \frac{-x}{2+x} - \left(\frac{-x}{1+x}\right) = \frac{-x(1+x) + x(2+x)}{(2+x)(1+x)} = \frac{-x - x^2 + 2x + x^2}{(2+x)(1+x)} = \frac{x}{(1+x)(2+x)}$.

   The equation becomes: $\sin^{-1}(A) = \frac{\pi}{2} - \cos^{-1}(B)$ Using the identity $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$, we have $\frac{\pi}{2} - \cos^{-1}(B) = \sin^{-1}(B)$. So, the equation simplifies to: $\sin^{-1}(A) = \sin^{-1}(B) \implies A = B$

3. Solve the algebraic equation $A=B$. $\frac{x^2}{(1-x)(2-x)} = \frac{x}{(1+x)(2+x)}$ One immediate solution is $x=0$. Let's check if it is valid. For $x=0$, $A=0$ and $B=0$. $\sin^{-1}(0) = \frac{\pi}{2} - \cos^{-1}(0) \implies 0 = \frac{\pi}{2} - \frac{\pi}{2} \implies 0=0$. So $x=0$ is a solution. It lies in the interval $(-1/2, 1/2)$.

   For $x \neq 0$, we can divide both sides by $x$: $\frac{x}{(1-x)(2-x)} = \frac{1}{(1+x)(2+x)}$ Cross-multiplying gives: $x(1+x)(2+x) = (1-x)(2-x)$ $x(x^2 + 3x + 2) = x^2 - 3x + 2$ $x^3 + 3x^2 + 2x = x^2 - 3x + 2$ $x^3 + 2x^2 + 5x - 2 = 0$

4. Find the number of roots of the cubic equation in $(-1/2, 1/2)$. Let $f(x) = x^3 + 2x^2 + 5x - 2$. We need to find the number of roots of $f(x)=0$ in the interval $(-1/2, 1/2)$.

   • Evaluate $f(x)$ at the interval boundaries: $f(-1/2) = (-1/2)^3 + 2(-1/2)^2 + 5(-1/2) - 2 = -1/8 + 1/2 - 5/2 - 2 = -1/8 - 4 = -33/8 < 0$. $f(1/2) = (1/2)^3 + 2(1/2)^2 + 5(1/2) - 2 = 1/8 + 1/2 + 5/2 - 2 = 1/8 + 3 - 2 = 1/8 + 1 = 9/8 > 0$. Since $f(x)$ is continuous and changes sign between $-1/2$ and $1/2$, by the Intermediate Value Theorem, there is at least one root in $(-1/2, 1/2)$.

   • Analyze the derivative of $f(x)$: $f'(x) = 3x^2 + 4x + 5$. The discriminant of this quadratic is $D = 4^2 - 4(3)(5) = 16 - 60 = -44 < 0$. Since the leading coefficient (3) is positive and the discriminant is negative, $f'(x)$ is always positive for all real $x$. This means $f(x)$ is a strictly increasing function.

   A strictly increasing function can cross the x-axis at most once. Therefore, there is exactly one root of $f(x)=0$ in the interval $(-1/2, 1/2)$. This root is not $x=0$ since $f(0) = -2 \neq 0$.

5. Check domain validity. The arguments of the inverse trigonometric functions, $A$ and $B$, must be in the range $[-1, 1]$. Since $A=B$, we only need to check one. Let's analyze $B(x) = \frac{x}{(1+x)(2+x)}$ for $x \in (-1/2, 1/2)$. The derivative is $B'(x) = \frac{(1+x)(2+x) - x(2x+3)}{((1+x)(2+x))^2} = \frac{2+3x+x^2 - 2x^2-3x}{(x^2+3x+2)^2} = \frac{2-x^2}{(x^2+3x+2)^2}$. For $x \in (-1/2, 1/2)$, $x^2 \in [0, 1/4)$, so $2-x^2 > 0$. Thus, $B'(x) > 0$ and $B(x)$ is strictly increasing on $(-1/2, 1/2)$. $B(-1/2) = \frac{-1/2}{(1/2)(3/2)} = -2/3$. $B(1/2) = \frac{1/2}{(3/2)(5/2)} = 2/15$. So for any $x \in (-1/2, 1/2)$, $B(x) \in (-2/3, 2/15)$, which is well within $[-1, 1]$. All solutions in the interval are valid.

6. Conclusion. We found two distinct real solutions in the interval $(-1/2, 1/2)$:

   1. $x=0$
   2. The unique real root of $x^3 + 2x^2 + 5x - 2 = 0$ that lies in $(-1/2, 1/2)$.

   Therefore, the total number of real solutions is 2.