If `cos^(-1)(x-(x^(2))/(2)+(x^(3))/(4)-…)" "+sin^(-1)(x^(2)-(x^(4))/(2)+(x^(6))/(4)-….)=(pi)/(2) 0le |x| lt sqrt2`, then x is equal to

Answer

To solve the equation

\[ \cos^{-1}\left(x - \frac{x^2}{2} + \frac{x^3}{4} - \ldots\right) + \sin^{-1}\left(x^2 - \frac{x^4}{2} + \frac{x^6}{4} - \ldots\right) = \frac{\pi}{2} \]

for \( 0 \leq |x| < \sqrt{2} \), we will first simplify the series in both the inverse cosine and inverse sine terms.

### Step 1: Simplifying the Cosine Inverse Term

The series in the cosine inverse term is:

\[ x - \frac{x^2}{2} + \frac{x^3}{4} - \ldots \]

This is a geometric series where the first term \( a = x \) and the common ratio \( r = -\frac{x}{2} \).

The sum of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} \]

Thus, we have:

\[ S = \frac{x}{1 - \left(-\frac{x}{2}\right)} = \frac{x}{1 + \frac{x}{2}} = \frac{x}{\frac{2 + x}{2}} = \frac{2x}{2 + x} \]

### Step 2: Simplifying the Sine Inverse Term

Now, we will simplify the sine inverse term:

\[ x^2 - \frac{x^4}{2} + \frac{x^6}{4} - \ldots \]

This is also a geometric series where the first term \( a = x^2 \) and the common ratio \( r = -\frac{x^2}{2} \).

Using the same formula for the sum of an infinite geometric series:

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

### Step 3: Substituting Back into the Equation

Now we substitute the simplified sums back into the original equation:

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

Using the property of inverse trigonometric functions:

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

we can equate the arguments:

\[ \frac{2x}{2 + x} = \frac{2x^2}{2 + x^2} \]

### Step 4: Cross-Multiplying

Cross-multiplying gives us:

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

Expanding both sides:

\[ 4x + 2x^3 = 4x^2 + 2x^3 \]

### Step 5: Simplifying the Equation

Subtract \( 2x^3 \) from both sides:

\[ 4x = 4x^2 \]

Dividing both sides by 4 (assuming \( x \neq 0 \)):

\[ x = x^2 \]

### Step 6: Solving the Quadratic Equation

Rearranging gives us:

\[ x^2 - x = 0 \]

Factoring out \( x \):

\[ x(x - 1) = 0 \]

Thus, the solutions are:

\[ x = 0 \quad \text{or} \quad x = 1 \]

### Conclusion

The values of \( x \) that satisfy the equation are:

\[ \boxed{0 \text{ and } 1} \]