Sum of roots of the equation `x^2-2x^2sin^2(pix)/2+1=0` is

Options

A. 0
B. 2
C. 1
D. 3

Explanation

To find the sum of the roots of the equation \( x^4 - 2x^2 \sin^2\left(\frac{\pi x}{2}\right) + 1 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^4 - 2x^2 \sin^2\left(\frac{\pi x}{2}\right) + 1 = 0 \]

### Step 2: Rearranging the equation
We can rearrange the equation to isolate the sine term:
\[ x^4 + 1 = 2x^2 \sin^2\left(\frac{\pi x}{2}\right) \]

### Step 3: Use the identity for squares
Notice that \( x^4 + 1 \) can be expressed as a difference of squares:
\[ x^4 + 1 = (x^2 - 1)^2 + 2x^2 \]
This means we can write:
\[ (x^2 - 1)^2 = 2x^2 \sin^2\left(\frac{\pi x}{2}\right) \]

### Step 4: Setting the expressions equal to zero
Now we can set each part of the equation to zero:
1. \( (x^2 - 1)^2 = 0 \)
2. \( 2x^2 \sin^2\left(\frac{\pi x}{2}\right) = 0 \)

### Step 5: Solve \( (x^2 - 1)^2 = 0 \)
From \( (x^2 - 1)^2 = 0 \), we find:
\[ x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1 \]

### Step 6: Solve \( 2x^2 \sin^2\left(\frac{\pi x}{2}\right) = 0 \)
For \( 2x^2 \sin^2\left(\frac{\pi x}{2}\right) = 0 \), we have two cases:
1. \( 2x^2 = 0 \implies x = 0 \)
2. \( \sin^2\left(\frac{\pi x}{2}\right) = 0 \implies \frac{\pi x}{2} = n\pi \) for \( n \in \mathbb{Z} \implies x = 2n \)

### Step 7: Collect all roots
The roots we have found are:
- From \( (x^2 - 1)^2 = 0 \): \( x = 1, -1 \)
- From \( 2x^2 = 0 \): \( x = 0 \)
- From \( \sin^2\left(\frac{\pi x}{2}\right) = 0 \): \( x = 2n \) (where \( n \) can be any integer, giving us roots like \( 2, -2, 4, -4, \ldots \))

### Step 8: Calculate the sum of the roots
The sum of the roots \( 1, -1, 0 \) is:
\[ 1 + (-1) + 0 = 0 \]
The other roots \( 2n \) do not contribute to the sum since they are symmetric around zero.

### Final Answer
Thus, the sum of the roots of the equation is:
\[ \text{Sum of roots} = 0 \]