Let
`Delta(theta)=|(1, sin theta, 1),(sin theta, 1, sin theta),(-1, -sin theta , 1)|, 0 le theta le 2pi`
solution of `Delta(theta)=3` is

Options

A. `{(pi)/2,(3pi)/2}`
B. `{(pi)/4,(3pi)/4,(5pi)/4,(7pi)/4}`
C. `{(pi)/4,(3pi)/4}`
D. `{(pi)/4,(pi)/2,(3pi)/4,pi}`

Answer

To solve the determinant \( \Delta(\theta) = \begin{vmatrix} 1 & \sin \theta & 1 \\ \sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix} \) and find the values of \( \theta \) for which \( \Delta(\theta) = 3 \), we will follow these steps:

### Step 1: Compute the Determinant
We will compute the determinant using the formula for a \( 3 \times 3 \) matrix:

\[ \Delta(\theta) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where:
- \( a = 1, b = \sin \theta, c = 1 \)
- \( d = \sin \theta, e = 1, f = \sin \theta \)
- \( g = -1, h = -\sin \theta, i = 1 \)

Substituting these values into the determinant formula:

\[ \Delta(\theta) = 1 \cdot (1 \cdot 1 - (-\sin \theta)(\sin \theta)) - \sin \theta \cdot (\sin \theta \cdot 1 - (-\sin \theta)(-1)) + 1 \cdot (\sin \theta(-\sin \theta) - 1 \cdot (-1)) \]

Calculating each term:

1. \( 1 \cdot (1 + \sin^2 \theta) = 1 + \sin^2 \theta \)
2. \( -\sin \theta \cdot (\sin \theta - \sin \theta) = 0 \)
3. \( 1 \cdot (-\sin^2 \theta + 1) = 1 - \sin^2 \theta \)

Combining these results:

\[ \Delta(\theta) = (1 + \sin^2 \theta) + (1 - \sin^2 \theta) = 2 \]

### Step 2: Set the Determinant Equal to 3
Now we set the determinant equal to 3:

\[ 2 = 3 \]

### Step 3: Analyze the Result
The equation \( 2 = 3 \) is a contradiction. This means that there are no values of \( \theta \) for which \( \Delta(\theta) = 3 \).

### Conclusion
Thus, the solution of \( \Delta(\theta) = 3 \) is that there are no solutions in the interval \( 0 \leq \theta \leq 2\pi \).

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