`|(sin^2x,cosx^2x,1), (cos^2x, sin^2x,1),(-10,12,2)|=0`

Options

A. 0
B. `12 cos^(2)x-10sin^(2)x`
C. `12sin^(2)x-10cos^(2)x-2`
D. `10sin2x`

Answer

To solve the determinant equation

\[ \begin{vmatrix} \sin^2 x & \cos^2 x & 1 \\ \cos^2 x & \sin^2 x & 1 \\ -10 & 12 & 2 \end{vmatrix} = 0, \]

we will follow these steps:

### Step 1: Write the determinant
We start with the determinant as given:

\[ D = \begin{vmatrix} \sin^2 x & \cos^2 x & 1 \\ \cos^2 x & \sin^2 x & 1 \\ -10 & 12 & 2 \end{vmatrix}. \]

### Step 2: Apply row operations
We can perform a row operation to simplify the determinant. Let's modify the first two rows by subtracting the first row from the second row:

\[ R_2 \rightarrow R_2 - R_1. \]

This gives us:

\[ D = \begin{vmatrix} \sin^2 x & \cos^2 x & 1 \\ \cos^2 x - \sin^2 x & \sin^2 x - \cos^2 x & 0 \\ -10 & 12 & 2 \end{vmatrix}. \]

### Step 3: Simplify the determinant
Now, we can simplify the determinant further. The second row now has a zero in the last column, which allows us to expand the determinant along the third column:

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

### Step 4: Calculate the 2x2 determinant
Now we compute the 2x2 determinant:

\[ D = \begin{vmatrix} \sin^2 x & \cos^2 x \\ \cos^2 x - \sin^2 x & \sin^2 x - \cos^2 x \end{vmatrix}. \]

Using the formula for the determinant of a 2x2 matrix, we have:

\[ D = \sin^2 x (\sin^2 x - \cos^2 x) - \cos^2 x (\cos^2 x - \sin^2 x). \]

### Step 5: Simplify further
This simplifies to:

\[ D = \sin^4 x - \sin^2 x \cos^2 x - \cos^4 x + \sin^2 x \cos^2 x. \]

The terms \(- \sin^2 x \cos^2 x\) and \(+\sin^2 x \cos^2 x\) cancel out, leaving us with:

\[ D = \sin^4 x - \cos^4 x. \]

### Step 6: Factor the expression
We can factor this as a difference of squares:

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

Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), we have:

\[ D = (\sin^2 x - \cos^2 x)(1). \]

### Step 7: Set the determinant to zero
For the determinant to equal zero, we need:

\[ \sin^2 x - \cos^2 x = 0. \]

This implies:

\[ \sin^2 x = \cos^2 x. \]

### Step 8: Solve for x
This occurs when:

\[ \tan^2 x = 1 \quad \Rightarrow \quad \tan x = \pm 1. \]

Thus, \(x = n\frac{\pi}{4} + k\pi\) for any integer \(n\).

### Conclusion
The solution to the determinant equation is:

\[ x = n\frac{\pi}{4} + k\pi \quad (n, k \in \mathbb{Z}). \]