Key Concepts and Formulas:

This problem primarily involves the properties of determinants and solving trigonometric equations.

1. Determinant Properties:
   • The value of a determinant remains unchanged if we apply row operations of the form $R_i \to R_i + kR_j$ (or column operations $C_i \to C_i + kC_j$).
   • If a row or column has a common factor, it can be taken out of the determinant.
   • If a row or column consists entirely of zeros, the determinant is zero.
   • If two rows or columns are identical, the determinant is zero.
   • The determinant of an upper or lower triangular matrix is the product of its diagonal elements.
2. Trigonometric Equations: Solving equations involving $\sin x$, $\cos x$, or $\tan x$ and finding solutions within a specified interval.

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Step-by-Step Solution:

We are given the equation: \left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0

Our goal is to find the values of $x$ in the interval $\left[ { - {\pi \over 4},{\pi \over 4}} \right]$ that satisfy this equation.

Step 1: Simplify the determinant using row operations. To simplify the determinant, we look for ways to create common factors or zeros. A common strategy for determinants where all elements are similar (like here, $\cos x$ on diagonal and $\sin x$ elsewhere) is to sum all rows/columns.

Apply the operation $R_1 \to R_1 + R_2 + R_3$: (This operation adds the elements of $R_2$ and $R_3$ to $R_1$, helping us find a common factor.) \left| {\matrix{ {\cos x + 2\sin x} & {\cos x + 2\sin x} & {\cos x + 2\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0

Now, we can factor out $(\cos x + 2\sin x)$ from the first row ($R_1$): (\cos x + 2\sin x) \left| {\matrix{ 1 & 1 & 1 \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0

Step 2: Further simplify the determinant using column operations. To make the determinant easier to expand, we can create more zeros in the first row.

Apply the operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$: (These operations subtract the first column from the second and third columns, creating zeros in the first row without changing the determinant's value.) (\cos x + 2\sin x) \left| {\matrix{ 1 & {1 - 1} & {1 - 1} \cr {\sin x} & {\cos x - \sin x} & {\sin x - \sin x} \cr {\sin x} & {\sin x - \sin x} & {\cos x - \sin x} \cr } } \right| = 0 (\cos x + 2\sin x) \left| {\matrix{ 1 & 0 & 0 \cr {\sin x} & {\cos x - \sin x} & 0 \cr {\sin x} & 0 & {\cos x - \sin x} \cr } } \right| = 0

Step 3: Expand the simplified determinant. The determinant is now in an upper triangular form (or can be seen as expanding along the first row). The determinant of a triangular matrix is the product of its diagonal elements. Expanding along the first row: $(\cos x + 2\sin x) \cdot \left[ 1 \cdot ((\cos x - \sin x)(\cos x - \sin x) - 0 \cdot 0) - 0 + 0 \right] = 0$ $(\cos x + 2\sin x) (\cos x - \sin x)^2 = 0$

Step 4: Solve the resulting trigonometric equations. For the product of terms to be zero, at least one of the terms must be zero. This gives us two possible cases:

Case 1: $\cos x - \sin x = 0$ $\cos x = \sin x$ Dividing by $\cos x$ (assuming $\cos x \neq 0$; if $\cos x = 0$, then $\sin x = \pm 1$, so $\sin x = \cos x$ would not hold. So $\cos x \neq 0$ is a valid assumption here): $\tan x = 1$

Case 2: $\cos x + 2\sin x = 0$ $\cos x = -2\sin x$ Dividing by $\cos x$ (again, $\cos x \neq 0$ for similar reasons): $1 = -2 \tan x$ $\tan x = - {1 \over 2}$

Step 5: Find the roots in the given interval $\left[ { - {\pi \over 4},{\pi \over 4}} \right]$.

For Case 1: $\tan x = 1$ The general solution for $\tan x = 1$ is $x = n\pi + {\pi \over 4}$, where $n$ is an integer. Let's check values of $n$:

• If $n=0$, $x = {\pi \over 4}$. This value is in the interval $\left[ { - {\pi \over 4},{\pi \over 4}} \right]$.
• If $n=1$, $x = \pi + {\pi \over 4} = {5\pi \over 4}$. This is outside the interval.
• If $n=-1$, $x = -\pi + {\pi \over 4} = -{3\pi \over 4}$. This is outside the interval. So, from Case 1, we have one distinct root: $x = {\pi \over 4}$.

For Case 2: $\tan x = - {1 \over 2}$ Let $\alpha = \arctan\left({1 \over 2}\right)$. Since $1/2$ is positive, $\alpha$ is in the first quadrant, i.e., $0 < \alpha < {\pi \over 2}$. Since $\tan x$ is negative, $x$ must be in the second or fourth quadrant. The general solution for $\tan x = - {1 \over 2}$ is $x = n\pi + \arctan\left(-{1 \over 2}\right)$, or $x = n\pi - \alpha$. Let's check values of $n$:

• If $n=0$, $x = \arctan\left(-{1 \over 2}\right)$. This value lies in the interval $\left( -{\pi \over 2}, 0 \right)$ because the principal value range of $\arctan$ is $\left( -{\pi \over 2}, {\pi \over 2} \right)$. To be precise, since $0 < {1 \over 2} < 1$, we know $0 < \arctan\left({1 \over 2}\right) < {\pi \over 4}$. Therefore, $x = -\arctan\left({1 \over 2}\right)$ is in the interval $\left( -{\pi \over 4}, 0 \right)$, which is within $\left[ -{\pi \over 4}, {\pi \over 4} \right]$.
• If $n=1$, $x = \pi - \arctan\left({1 \over 2}\right)$. This is clearly outside the interval.
• If $n=-1$, $x = -\pi - \arctan\left({1 \over 2}\right)$. This is also outside the interval. So, from Case 2, we have one distinct root: $x = -\arctan\left({1 \over 2}\right)$.

Step 6: Count the distinct real roots. The distinct real roots in the given interval are:

1. $x = {\pi \over 4}$
2. $x = -\arctan\left({1 \over 2}\right)$

Both values are distinct and lie within the specified interval $\left[ { - {\pi \over 4},{\pi \over 4}} \right]$. Therefore, there are 2 distinct real roots.

The provided solution states the answer is (A) 4. Let's re-check the question and solution carefully. Ah, the provided correct answer is (A) 4, but my derived answer is 2. Let's check the original solution again. Original solution: Expanding using first column, (cosx - sinx)(cos - sinx) (sinx + cos x) + sinx (cosx - sinx) (sinx - cosx) = 0 => (cosx - sinx) 2 (sinx + cosx) + sinx (cosx - sinx) 2 = 0 => (cosx - sinx) 2 (sinx + cosx + sinx) = 0 => (2sinx + cosx )(cosx - sinx) 2 = 0

This factorization is identical to what I derived: $(\cos x + 2\sin x) (\cos x - \sin x)^2 = 0$. The roots are $\tan x = 1$ and $\tan x = -1/2$. My