The number of distinct real roots of `abs((sinx, cosx, cosx),(cos x,sin x,cos x),(cos x,cos x,sin x))=0` in the interval `-(pi)/4 le x le (pi)/4` is

Options

A. 0
B. 2
C. 1
D. 3

Explanation

To find the number of distinct real roots of the determinant

\[ D = \begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} \]

in the interval \(-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}\), we will compute the determinant and set it equal to zero.

### Step 1: Calculate the Determinant

Using the formula for the determinant of a 3x3 matrix, we have:

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

where the matrix is:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

For our matrix:

- \(a = \sin x\), \(b = \cos x\), \(c = \cos x\)
- \(d = \cos x\), \(e = \sin x\), \(f = \cos x\)
- \(g = \cos x\), \(h = \cos x\), \(i = \sin x\)

Calculating each part, we have:

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

This simplifies to:

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

### Step 2: Simplify the Expression

Notice that \(\sin^2 x - \cos^2 x = -\cos(2x)\). The determinant can be rewritten as:

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

Combining like terms and simplifying yields:

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

### Step 3: Set the Determinant Equal to Zero

To find the roots, we set the determinant equal to zero:

\[ -\sin x \cos(2x) = 0 \]

This gives us two cases to consider:

1. \(\sin x = 0\)
2. \(\cos(2x) = 0\)

### Step 4: Solve Each Case

**Case 1: \(\sin x = 0\)**

In the interval \(-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}\), \(\sin x = 0\) at:

\[ x = 0 \]

**Case 2: \(\cos(2x) = 0\)**

The equation \(\cos(2x) = 0\) gives:

\[ 2x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

This simplifies to:

\[ x = \frac{\pi}{4} + \frac{n\pi}{2} \]

In the interval \(-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}\), the only solution is:

\[ x = \frac{\pi}{4} \]

### Step 5: Count the Distinct Real Roots

Thus, the distinct real roots in the interval \(-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}\) are:

1. \(x = 0\)
2. \(x = \frac{\pi}{4}\)

This gives us a total of **2 distinct real roots**.

### Final Answer
The number of distinct real roots of the determinant in the given interval is **2**.