Q.The number of distinct real roots of $\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$ in the interval $-\dfrac{\pi}{4} \le x \le \dfrac{\pi}{4}$ is
(A) $0$
(B) $2$
(C) $1$
(D) $3$

Concept: Determinant Equality Equation — We simplify the determinant using row/column operations, then solve the resulting trigonometric equation within the given interval.

Step 1: Simplify the determinant.

Let $C_1 \to C_1 + C_2 + C_3$. Each entry in $C_1$ becomes $\sin x + 2\cos x$. Factor this common term out:

Step 2: Subtract $R_1$ from $R_2$ and $R_3$.

The determinant becomes:

This is upper triangular, so the determinant is $(\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$.

Step 3: Solve in $[-\pi/4, \pi/4]$. …

We are solving a determinant equality equation — a matrix whose entries are trigonometric functions set to zero. The key is to simplify the determinant into a product of factors, each a simple trigonometric equation. Then we check which of those equations have solutions inside the narrow interval $-\pi/4 \le x \le \pi/4$.

The matrix is symmetric and has a special pattern: all diagonal entries are $\sin x$, all off-diagonal entries are $\cos x$. This is a classic “all-entries-equal-off-diagonal” matrix, which can be handled by adding rows or columns, or by using the eigenvalue approach.

1. Simplify the determinant using row operations. Let

Add all three rows to the first row. That is, $R_1 \to R_1 + R_2 + R_3$.

The first row becomes:

and the same for each column, so the new first row is:

The determinant becomes:

2. Factor out the common factor from the first row. Since every entry in row 1 has the factor $(\sin x + 2\cos x)$, we pull it out:

3. Simplify the remaining $3\times3$ determinant. Subtract column 1 from columns 2 and 3: $C_2 \to C_2 - C_1$, $C_3 \to C_3 - C_1$. The determinant becomes:

This is now upper triangular (in fact, diagonal after the first row). The value is the product of the diagonal entries:

4. Thus the full determinant is:

Setting $\Delta = 0$ gives two families of equations:

$\sin x + 2\cos x = 0 \quad \text{or} \quad \sin x - \cos x = 0$ …

Method: Solving a Trigonometric Determinant Equation on a Restricted Interval

When a determinant with a repeated symmetric trigonometric pattern is set to zero, simplify it into a product of simple trigonometric factors, then solve each factor only within the interval the question actually asks about.

Steps

Step 1: Expose a common factor via a row/column sum

Use an operation like $C_1 \to C_1+C_2+C_3$ on a matrix with repeated entries to make a whole row/column collapse to a single common trigonometric expression, then factor it out.

Step 2: Reduce what remains to triangular form

Subtract one row/column from the others to create zeros, so the remaining determinant becomes the product of its diagonal entries — this gives the full determinant as a product of simple trigonometric factors.

Step 3: Solve each factor as its own equation …

Common Mistakes

Mistake 1: Forgetting to check whether $\tan x=-2$'s solution actually lies in the interval

Why it's wrong: $\arctan(-2)\approx-63.4^\circ\approx-1.107$ rad, which is well outside $[-\pi/4,\pi/4]\approx[-0.785,0.785]$; a student who solves $\tan x=-2$ and assumes it automatically contributes a root will overcount and pick option (B) $2$ or (D) $3$ instead of (C). Correct approach: convert every candidate angle to a numeric value and explicitly compare it against the interval's numeric bounds before counting it as a valid root.

Mistake 2: Counting the repeated factor $(\sin x-\cos x)^2$ as two distinct roots …