The number of distinct real roots of | sin x cos x cos x | | cos x sin x cos x | | cos x cos x sin x

Q: Why can't I just expand the determinant using Sarrus' rule or cofactor expansion?

You *could*, but you shouldn't. In JEE Advanced, time is your most precious currency. Expanding directly leads to a massive expression that is prone to algebraic errors. The symmetry is a signal—a hint from the examiner—to use row operations to simplify the structure first. Always look for patterns before you start calculating.

Q: Wait, when we solve $\sin x + 2\cos x = 0$, we divide by $\cos x$. Is it possible that $\cos x = 0$?

Excellent observation! Always check for division by zero. In our interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$, the value of $\cos x$ ranges from $\frac{1}{\sqrt{2}}$ to $1$. Since $\cos x$ is never zero in this interval, dividing by it is perfectly safe.

Q: Why do we reject $\tan x = -2$?

Because of the boundary conditions! We are restricted to $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$. In this specific window, the function $y = \tan x$ only takes values between $-1$ and $1$. Since $-2$ is outside this range, it is impossible for $\tan x$ to equal $-2$ within our domain.

Question

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 $-\pi/4 \le x \le \pi/4$ is

Options:

(A) $0$

(B) $2$

(C) $1$

(D) $3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a determinant; we are learning to see the hidden architecture of mathematics. When you look at the matrix

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

If your instinct is to immediately start expanding along the first row, pause. In the world of JEE Advanced, brute force is rarely the intended path. The problem is designed with a beautiful, rhythmic symmetry where every row is a permutation of the others.

### The Power of Row Operations

Let us apply the most elegant tool in our arsenal: row operations. We want to simplify the matrix while preserving its value. Consider the operation $R_1 	o R_1 + R_2 + R_3$.

By summing all rows into the first row, we transform the top line into a uniform sequence of identical terms: $(\sin x + 2\cos x)$. The determinant becomes:

$$\begin{vmatrix} \sin x + 2\cos x & \sin x + 2\cos x & \sin x + 2\cos x \ \cos x & \sin x & \cos x \ \cos x & \cos x & \sin x \end{vmatrix} = 0$$

We can pull that common factor, $(\sin x + 2\cos x)$, completely out of the determinant. This is the moment the problem breaks open, leaving us with a row of ones—the ultimate invitation to create zeros.

### Creating Zeros: The Simplification

With a row of ones, our next move is clear. We apply $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$ to reduce the matrix to a much friendlier form:

$$(\sin x + 2\cos x) \begin{vmatrix} 1 & 0 & 0 \ \cos x & \sin x - \cos x & 0 \ \cos x & 0 & \sin x - \cos x \end{vmatrix} = 0$$

Expanding this is now a trivial task. We are left with the elegant product:

$$(\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$$

This is the heart of the problem. We have reduced a complex trigonometric determinant into two simple algebraic cases.

### The Final Verdict: Respecting the Domain

Now, we must be careful. We have two cases:

1. $\sin x + 2\cos x = 0 \implies 	an x = -2$
2. $(\sin x - \cos x)^2 = 0 \implies 	an x = 1$

We are working within the strict interval $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$. If you visualize the function $y = 	an x$ on this interval, you will see that it is strictly increasing from $-1$ to $1$.

Look at our first case: $	an x = -2$. Since $-2$ is less than $-1$, this line never intersects our curve within the allowed region. It is an extraneous solution.

Look at our second case: $	an x = 1$. This occurs exactly at the boundary $x = \frac{\pi}{4}$. This is a valid root.

Therefore, despite having two algebraic solutions, only one survives the test of the domain. The **number of distinct real roots is exactly 1**. Trust the process, respect the domain, and always verify your results.

Animated Solution for Mathematics - Matrices and Determinants: The number of distinct real roots of $sin x cos x cos x cos x sin x cos x cos x cos x sin x = 0$ in the interval $- π /4 \leq x \leq π /4$ is

Select Answer:

Visualized Solution

Analyze the Determinant Structure

Applying Row Operations

Extracting the Common Factor

Creating Zeros via Column Operations

Expanding the Determinant

Solving the First Factor

Solving the Second Factor

Visualizing the Interval

Analyzing the Range of $tan x$

Checking $tan x = - 2$

Checking $tan x = 1$

Final Conclusion

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Power of Row Operations

Creating Zeros

The Final Verdict

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