The general value of x which satisfies the equation `( cos x + i sin x) ( cos 3 x + i sin 3 x)`
` ( cos 5 x + i sin 5 x) . . . [ cos ( 2 n - 1) x + i sin ( 2 n - 1) x] = 1` is

Options

A. `(r pi)/( n^(2))`
B. `(( r - 1) pi)/( n^(2))`
C. `((2 r + 1) pi)/( n^(3))`
D. ` (2 r pi)/( n^(2))`

Answer

To solve the equation

\[ (\cos x + i \sin x)(\cos 3x + i \sin 3x)(\cos 5x + i \sin 5x) \cdots (\cos(2n-1)x + i \sin(2n-1)x) = 1, \]

we can utilize the fact that \( \cos \theta + i \sin \theta = e^{i\theta} \). Thus, we can rewrite the left-hand side as:

\[ e^{ix} \cdot e^{i3x} \cdot e^{i5x} \cdots e^{i(2n-1)x} = e^{i(x + 3x + 5x + \cdots + (2n-1)x)}. \]

### Step 1: Summing the angles

The sum of the angles can be expressed as:

\[ x + 3x + 5x + \cdots + (2n-1)x = x(1 + 3 + 5 + \cdots + (2n-1)). \]

The series \(1 + 3 + 5 + \cdots + (2n-1)\) is the sum of the first \(n\) odd numbers, which can be calculated using the formula:

\[ \text{Sum} = n^2. \]

Thus, we have:

\[ x(1 + 3 + 5 + \cdots + (2n-1)) = x \cdot n^2. \]

### Step 2: Rewriting the equation

Now we can rewrite the equation as:

\[ e^{i n^2 x} = 1. \]

### Step 3: Finding the general solution

The expression \( e^{i n^2 x} = 1 \) holds true when:

\[ n^2 x = 2r\pi, \quad r \in \mathbb{Z}. \]

### Step 4: Solving for \(x\)

From this, we can solve for \(x\):

\[ x = \frac{2r\pi}{n^2}, \quad r \in \mathbb{Z}. \]

### Final Answer

Thus, the general value of \(x\) that satisfies the given equation is:

\[ x = \frac{2r\pi}{n^2}, \quad r \in \mathbb{Z}. \]

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