The value of the determinants `|{:(1,a,a^(2)),(cos(n-1)x,cos nx , cos(n+1)x),(sin(n-1)x , sin nx , sin(n+1)x):}|` is zero if

Options

A. ` x = n pi `
B. `x = n pi //2 `
C. `x=(2n+1)pi//2`
D. `x=(1+a^(2))/(2a)n in I `

Answer

To solve the determinant

\[ D = \begin{vmatrix} 1 & a & a^2 \\ \cos((n-1)x) & \cos(nx) & \cos((n+1)x) \\ \sin((n-1)x) & \sin(nx) & \sin((n+1)x) \end{vmatrix} \]

we need to find the values of \(x\) for which \(D = 0\).

### Step 1: Substitute \(n = 1\)

To simplify the determinant, we can set \(n = 1\):

\[ D = \begin{vmatrix} 1 & a & a^2 \\ \cos(0) & \cos(x) & \cos(2x) \\ \sin(0) & \sin(x) & \sin(2x) \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} 1 & a & a^2 \\ 1 & \cos(x) & \cos(2x) \\ 0 & \sin(x) & \sin(2x) \end{vmatrix} \]

### Step 2: Use the identity for \(\sin(2x)\)

Using the identity \(\sin(2x) = 2\sin(x)\cos(x)\), we can rewrite the determinant:

\[ D = \begin{vmatrix} 1 & a & a^2 \\ 1 & \cos(x) & \cos(2x) \\ 0 & \sin(x) & 2\sin(x)\cos(x) \end{vmatrix} \]

### Step 3: Factor out \(\sin(x)\)

We can factor \(\sin(x)\) from the third row:

\[ D = \sin(x) \begin{vmatrix} 1 & a & a^2 \\ 1 & \cos(x) & \cos(2x) \\ 0 & 1 & 2\cos(x) \end{vmatrix} \]

### Step 4: Expand the determinant

Now we expand the determinant:

\[ D = \sin(x) \left[ 1 \cdot \begin{vmatrix} \cos(x) & \cos(2x) \\ 1 & 2\cos(x) \end{vmatrix} - a \cdot \begin{vmatrix} 1 & \cos(2x) \\ 0 & 2\cos(x) \end{vmatrix} + a^2 \cdot \begin{vmatrix} 1 & \cos(x) \\ 0 & 1 \end{vmatrix} \right] \]

Calculating the 2x2 determinants, we get:

\[ D = \sin(x) \left[ \cos(x) \cdot (2\cos(x)) - \cos(2x) - a \cdot (2\cos(x)) + a^2 \right] \]

### Step 5: Set the determinant to zero

For the determinant to be zero, we have two cases:

1. \(\sin(x) = 0\)
2. The remaining determinant equals zero.

### Case 1: Solve \(\sin(x) = 0\)

\(\sin(x) = 0\) gives us:

\[ x = n\pi \quad (n \in \mathbb{Z}) \]

### Case 2: Solve the remaining determinant

The remaining determinant simplifies to:

\[ 2\cos^2(x) - \cos(2x) - 2a\cos(x) + a^2 = 0 \]

Using the identity \(\cos(2x) = 2\cos^2(x) - 1\), we substitute:

\[ 2\cos^2(x) - (2\cos^2(x) - 1) - 2a\cos(x) + a^2 = 0 \]

This simplifies to:

\[ 1 + a^2 - 2a\cos(x) = 0 \]

Rearranging gives:

\[ 2a\cos(x) = 1 + a^2 \quad \Rightarrow \quad \cos(x) = \frac{1 + a^2}{2a} \]

### Step 6: Find \(x\)

Thus, we have:

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

### Final Solutions

The values of \(x\) for which the determinant is zero are:

1. \(x = n\pi\) for integers \(n\)
2. \(x = \cos^{-1}\left(\frac{1 + a^2}{2a}\right)\)