If `Sigma_(i=1)^(2n) cos^(-1) x_(i) = 0`,then find the value of ` Sigma_(i=1)^(2n) x_(i)`

Answer

To solve the problem, we need to analyze the given equation:

\[ \Sigma_{i=1}^{2n} \cos^{-1}(x_i) = 0 \]

### Step 1: Understanding the Range of \(\cos^{-1}(x)\)
The function \(\cos^{-1}(x)\) (inverse cosine) has a range of \([0, \pi]\). This means that for any real number \(x\) in the domain of \([-1, 1]\), \(\cos^{-1}(x)\) will always yield a non-negative value.

**Hint:** Recall that the output of \(\cos^{-1}(x)\) is always non-negative.

### Step 2: Analyzing the Summation
Since the sum of several non-negative terms equals zero, each individual term must also be zero. Therefore, we can conclude that:

\[ \cos^{-1}(x_i) = 0 \quad \text{for all } i \]

**Hint:** If the sum of non-negative numbers is zero, each number must be zero.

### Step 3: Finding \(x_i\)
From the equation \(\cos^{-1}(x_i) = 0\), we can find the value of \(x_i\):

\[ x_i = \cos(0) = 1 \]

This means that for all \(i\) from \(1\) to \(2n\):

\[ x_1 = x_2 = x_3 = \ldots = x_{2n} = 1 \]

**Hint:** Use the property of the cosine function to find the value of \(x_i\).

### Step 4: Calculating the Summation of \(x_i\)
Now, we need to find the summation:

\[ \Sigma_{i=1}^{2n} x_i = x_1 + x_2 + x_3 + \ldots + x_{2n} \]

Since all \(x_i\) are equal to \(1\):

\[ \Sigma_{i=1}^{2n} x_i = 1 + 1 + 1 + \ldots + 1 \quad (2n \text{ times}) \]

This simplifies to:

\[ \Sigma_{i=1}^{2n} x_i = 2n \]

### Final Answer
Thus, the value of \(\Sigma_{i=1}^{2n} x_i\) is:

\[ \boxed{2n} \]