The value of x that satisfies the relation `x=1 -x+x^2-x^3+x^4-x^5+….oo is `

Options

A. `2 cos 36^@`
B. `2 cos 144^@`
C. `2 sin 18^@`
D. `2 cos 18^@`

Explanation

To solve the equation \( x = 1 - x + x^2 - x^3 + x^4 - x^5 + \ldots \), we can recognize that the right-hand side is an infinite geometric series. Let's break down the steps to find the value of \( x \).

### Step 1: Identify the Infinite Series
The series can be expressed as:
\[ S = 1 - x + x^2 - x^3 + x^4 - x^5 + \ldots \]
This is an infinite geometric series where the first term \( a = 1 \) and the common ratio \( r = -x \).

### Step 2: Sum of the Infinite Series
The sum \( S \) of an infinite geometric series is given by the formula:
\[ S = \frac{a}{1 - r} \]
provided that \( |r| < 1 \). In our case:
\[ S = \frac{1}{1 - (-x)} = \frac{1}{1 + x} \]

### Step 3: Set Up the Equation
Now we can substitute this back into our equation:
\[ x = \frac{1}{1 + x} \]

### Step 4: Clear the Denominator
To eliminate the fraction, multiply both sides by \( 1 + x \):
\[ x(1 + x) = 1 \]
This simplifies to:
\[ x + x^2 = 1 \]

### Step 5: Rearrange the Equation
Rearranging gives us a standard quadratic equation:
\[ x^2 + x - 1 = 0 \]

### Step 6: Apply the Quadratic Formula
We can solve for \( x \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = -1 \). Plugging in these values:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 7: Determine Valid Roots
This gives us two potential solutions:
\[ x_1 = \frac{-1 - \sqrt{5}}{2} \quad \text{and} \quad x_2 = \frac{-1 + \sqrt{5}}{2} \]
Since \( x \) must be non-negative for the infinite series to converge, we reject \( x_1 \) because it is negative.

### Step 8: Final Result
Thus, the valid solution is:
\[ x = \frac{-1 + \sqrt{5}}{2} \]
This can also be expressed as:
\[ x = 2 \sin 18^\circ \]

### Conclusion
The value of \( x \) that satisfies the relation is:
\[ \boxed{2 \sin 18^\circ} \]