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 i

Q: Why can't I just expand the determinant directly? It's only a 3x3 matrix.

You absolutely could, but in the JEE Advanced, time is your most precious currency. Expanding a 3x3 determinant with trigonometric functions directly leads to a massive, messy expression that is prone to sign errors. By using row operations, we reduce the complexity before we even begin the algebra. It is about working smarter, not harder.

Q: How do I know which row operations to choose? Is there a rule?

Great question! The rule is 'Look for the Sum.' Notice that if you add all columns or all rows, you get the same expression $(\sin x + 2\cos x)$. Alternatively, subtracting rows creates zeros and common factors. When you see a matrix where elements are repeated across rows, always try to create a row of zeros or a row with identical terms. It is a classic pattern in determinant problems.

Q: Why is $\tan x = -2$ rejected? I solved it correctly!

You solved the algebra correctly, but you missed the 'Gatekeeper'—the domain constraint. We are restricted to $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$. In this interval, the function $\tan x$ ranges only from $-1$ to $1$. Since $-2$ is outside this range, the equation $\tan x = -2$ has no solution in our specific window. Always check your domain before finalizing your answer!

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 $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$ is:

Options:

(A) 4

(B) 1

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of trigonometric expansion.

We are faced with a $3 	imes 3$ determinant. The instinct for many students is to dive headfirst into the standard expansion formula. But wait—stop.

In the world of JEE Advanced, whenever you see a determinant with repeating elements like $\sin x$ and $\cos x$ arranged in this specific, symmetric pattern, you are not looking at a calculation problem; you are looking at a puzzle of symmetry.

### The Surgical Strike: Row Operations

Our objective is to simplify. We want to transform this matrix into something manageable. We look at the rows:

$$R_1 = [\sin x, \cos x, \cos x]$$
$$R_2 = [\cos x, \sin x, \cos x]$$
$$R_3 = [\cos x, \cos x, \sin x]$$

Notice the beauty here? If we subtract $R_2$ from $R_1$, we get $(\sin x - \cos x)$ in the first column and $(\cos x - \sin x)$ in the second. That is a common factor waiting to be pulled out!

Let us perform the operations $R_1 	o R_1 - R_2$ and $R_2 	o R_2 - R_3$.

Suddenly, the matrix transforms. The first row becomes $(\sin x - \cos x, \cos x - \sin x, 0)$. The second row becomes $(0, \sin x - \cos x, \cos x - \sin x)$.

We have created zeros and a common factor of $(\sin x - \cos x)$. By factoring this out from both rows, we have effectively performed 'mathematical surgery,' removing the complexity and leaving behind the core structure.

### The Factored Elegance

After pulling out the common terms, we are left with the equation:

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

Now, expanding this $3 	imes 3$ determinant is a trivial task. Expanding along the first row, we get:

$$1(\sin x - (-\cos x)) - (-1)(0 - (-\cos x)) + 0$$

This simplifies beautifully to $(\sin x + \cos x) - \cos x$, which is simply $\sin x$. Wait—let us re-evaluate the expansion carefully. The determinant evaluates to:

$$1(\sin x + \cos x) + 1(0 + \cos x) = \sin x + 2\cos x$$

Look at what we have achieved! The entire, intimidating determinant has collapsed into a neat, elegant product:

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

### The Final Gatekeeper: Domain Constraints

We have two distinct cases to solve.

Case 1: $(\sin x - \cos x)^2 = 0$, which implies $\sin x = \cos x$, or $	an x = 1$.

Case 2: $\sin x + 2\cos x = 0$, which implies $\sin x = -2\cos x$, or $	an x = -2$.

Now, we must respect the interval constraint: $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$. This is where many students stumble.

In our interval $[-\frac{\pi}{4}, \frac{\pi}{4}]$, the function $y = 	an x$ is strictly increasing from $-1$ to $1$.

For Case 1, $	an x = 1$. This intersects our curve exactly at the boundary $x = \frac{\pi}{4}$. That is one valid root.

For Case 2, $	an x = -2$. Look at the graph again. The lowest value our function reaches is $-1$. The line $y = -2$ is completely below our curve and never touches it.

### Conclusion

We started with a complex determinant, used the power of row operations to simplify it, factored it into a beautiful product, and finally used graphical intuition to navigate the domain constraints.

We found exactly **one root**. This is the essence of JEE Advanced mathematics—not just brute force, but the elegant application of logic and visualization. You have mastered this problem.

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 $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is:

Select Answer:

Visualized Solution

The Determinant Equation

Row Operations Strategy

Applying $R_{1} \to R_{1} - R_{2}$

Applying $R_{2} \to R_{2} - R_{3}$

Factoring Out Common Terms

Expanding the Determinant

The Factored Equation

Splitting into Cases

Visualizing the Domain

Analyzing Case 1: $tan x = 1$

Analyzing Case 2: $tan x = - 2$

Final Count of Roots

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Surgical Strike

The Factored Elegance

The Final Gatekeeper

Conclusion

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