The value of `theta` lying between 0 and `pi/2` and satisfying `|[1+sin^2theta,cos^2theta,4sin4theta],[sin^2theta,1+cos^2theta,4sin4theta],[sin^2theta,cos^2theta,1+4sin4theta]|=0`

Options

A. `(7pi)/(24)and (11pi)/(24)`
B. `(7pi)/(24)and (5pi)/(24)`
C. `(5pi)/(24)and (pi)/(24)`
D. none of these

Answer

To solve the problem, we need to find the value of \(\theta\) that satisfies the determinant condition given in the question. The determinant of the matrix is set to zero, and we will solve it step by step.

### Step 1: Write down the determinant
We start with the determinant:
\[ D = \left| \begin{array}{ccc} 1 + \sin^2 \theta & \cos^2 \theta & 4 \sin 4\theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4\theta \end{array} \right| = 0 \]

### Step 2: Apply row operations
We can simplify the determinant by performing row operations. Subtract the second row from the first row and the third row from the second row:
\[ D = \left| \begin{array}{ccc} 1 + \sin^2 \theta - \sin^2 \theta & \cos^2 \theta - (1 + \cos^2 \theta) & 4 \sin 4\theta - 4 \sin 4\theta \\ \sin^2 \theta - \sin^2 \theta & 1 + \cos^2 \theta - \cos^2 \theta & 4 \sin 4\theta - 4 \sin 4\theta \\ \sin^2 \theta - \sin^2 \theta & \cos^2 \theta - \cos^2 \theta & 1 + 4 \sin 4\theta - (1 + \cos^2 \theta) \end{array} \right| \]
This simplifies to:
\[ D = \left| \begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \sin 4\theta - 1 \end{array} \right| \]

### Step 3: Calculate the determinant
The determinant of this upper triangular matrix is the product of the diagonal elements:
\[ D = 1 \cdot 1 \cdot (4 \sin 4\theta - 1) = 4 \sin 4\theta - 1 \]

### Step 4: Set the determinant to zero
Now we set the determinant equal to zero:
\[ 4 \sin 4\theta - 1 = 0 \]
This gives:
\[ 4 \sin 4\theta = 1 \quad \Rightarrow \quad \sin 4\theta = \frac{1}{4} \]

### Step 5: Solve for \(4\theta\)
To find \(4\theta\), we take the inverse sine:
\[ 4\theta = \arcsin\left(\frac{1}{4}\right) + 2k\pi \quad \text{or} \quad 4\theta = \pi - \arcsin\left(\frac{1}{4}\right) + 2k\pi \]
for \(k \in \mathbb{Z}\).

### Step 6: Find \(\theta\)
Dividing by 4 gives:
\[ \theta = \frac{1}{4} \arcsin\left(\frac{1}{4}\right) + \frac{k\pi}{2} \quad \text{or} \quad \theta = \frac{\pi}{4} - \frac{1}{4} \arcsin\left(\frac{1}{4}\right) + \frac{k\pi}{2} \]

### Step 7: Determine valid \(\theta\) in \((0, \frac{\pi}{2})\)
We need to find values of \(\theta\) that lie within the interval \((0, \frac{\pi}{2})\).

1. For \(k = 0\):
- \(\theta = \frac{1}{4} \arcsin\left(\frac{1}{4}\right)\) is valid.
- \(\theta = \frac{\pi}{4} - \frac{1}{4} \arcsin\left(\frac{1}{4}\right)\) is also valid.

2. For \(k = 1\) or higher, \(\theta\) will exceed \(\frac{\pi}{2}\).

### Conclusion
Thus, the values of \(\theta\) that satisfy the given determinant condition and lie between \(0\) and \(\frac{\pi}{2}\) are:
\[ \theta = \frac{1}{4} \arcsin\left(\frac{1}{4}\right) \quad \text{and} \quad \theta = \frac{\pi}{4} - \frac{1}{4} \arcsin\left(\frac{1}{4}\right) \]