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

Options

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

Explanation

To find the number of distinct real roots of the determinant equation

\[ \begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0 \]

in the interval \(-\frac{\pi}{4} \leq x \leq t \frac{\pi}{4}\), we will follow these steps:

### Step 1: Calculate the Determinant
We start by calculating the determinant of the matrix:

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

Using the determinant formula for a \(3 \times 3\) matrix, we can expand it:

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

Calculating the \(2 \times 2\) determinants:

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

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

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

Substituting these back into the determinant:

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

### Step 2: Simplify the Expression
Now we simplify \(D\):

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

Combining like terms and simplifying will yield a polynomial in terms of \(\sin x\) and \(\cos x\).

### Step 3: Set the Determinant to Zero
We set \(D = 0\) and solve for \(x\):

\[ \sin x + 2\cos x = 0 \quad \text{and} \quad \sin x - \cos x = 0 \]

From \(\sin x + 2\cos x = 0\):

\[ \tan x = -2 \]

From \(\sin x - \cos x = 0\):

\[ \tan x = 1 \]

### Step 4: Find the Roots
1. For \(\tan x = -2\):
- The general solutions are \(x = \tan^{-1}(-2) + n\pi\).

2. For \(\tan x = 1\):
- The general solutions are \(x = \frac{\pi}{4} + n\pi\).

### Step 5: Determine Roots in the Given Interval
Now we need to find how many of these roots fall within the interval \(-\frac{\pi}{4} \leq x \leq t \frac{\pi}{4}\).

1. For \(\tan x = -2\):
- The principal value is \(x_1 = \tan^{-1}(-2)\), which is approximately \(-1.107\) (in radians).
- This root is within the interval.

2. For \(\tan x = 1\):
- The principal value is \(x_2 = \frac{\pi}{4}\), which is also within the interval.

### Conclusion
Thus, we have two distinct roots in the interval \(-\frac{\pi}{4} \leq x \leq t \frac{\pi}{4}\).

### Final Answer
The number of distinct real roots is **2**.

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