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

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

and find the condition under which \(D = 0\), we can follow these steps:

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

Assuming \(n = 1\), the determinant simplifies to:

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

This simplifies to:

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

### Step 2: Expand the Determinant

We can expand the determinant along the first row:

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

Calculating these 2x2 determinants:

1. \(\begin{vmatrix} \cos(x) & \cos(2x) \\ \sin(x) & \sin(2x) \end{vmatrix} = \cos(x) \sin(2x) - \sin(x) \cos(2x) = \sin(2x - x) = \sin(x)\)

2. \(\begin{vmatrix} 1 & \cos(2x) \\ 0 & \sin(2x) \end{vmatrix} = 1 \cdot \sin(2x) - 0 = \sin(2x)\)

3. \(\begin{vmatrix} 1 & \cos(x) \\ 0 & \sin(x) \end{vmatrix} = 1 \cdot \sin(x) - 0 = \sin(x)\)

Putting it all together, we have:

\[ D = \sin(x) - a \sin(2x) + a^2 \sin(x) \]

### Step 3: Factor the Expression

Now we can factor out \(\sin(x)\):

\[ D = \sin(x)(1 + a^2 - a \cdot 2 \cos(x)) \]

### Step 4: Set the Determinant to Zero

For the determinant to be zero, we need:

\[ \sin(x) = 0 \quad \text{or} \quad 1 + a^2 - 2a \cos(x) = 0 \]

### Step 5: Solve for \(x\)

1. **From \(\sin(x) = 0\)**:
\[ x = n\pi, \quad n \in \mathbb{Z} \]

2. **From \(1 + a^2 - 2a \cos(x) = 0\)**:
Rearranging gives:
\[ 2a \cos(x) = 1 + a^2 \quad \Rightarrow \quad \cos(x) = \frac{1 + a^2}{2a} \]
Thus, \(x = \cos^{-1}\left(\frac{1 + a^2}{2a}\right)\).

### Conclusion

The value of the determinant is zero if:

\[ x = n\pi \quad \text{or} \quad \cos(x) = \frac{1 + a^2}{2a} \]