The values of θ lying between θ = 0 and θ = π/2 and satisfying the equation | 1 + sin² θ cos² θ 4 si

Q: Why can't we just expand the determinant directly? It seems like a standard procedure.

You absolutely could, but it would be a nightmare of algebraic terms! In JEE Advanced, the goal is often to find the most elegant path. By using row operations, we reduce the complexity of the determinant before expanding, which prevents calculation errors and saves precious time.

Q: Why did the range of $4\theta$ change from $(0, \pi/2)$ to $(0, 2\pi)$?

Great catch! The original constraint is for $\theta$, which is $0 < \theta < \frac{\pi}{2}$. Since our equation involves $\sin 4\theta$, we must consider the range of the argument $4\theta$. Multiplying the entire inequality by $4$ gives $4 \times 0 < 4\theta < 4 \times \frac{\pi}{2}$, which is $0 < 4\theta < 2\pi$. Always check your domain!

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

There is no single rule, but there is a strategy: look for rows that are almost identical. In this problem, the rows were very similar, differing only by small constants or terms. Subtracting one row from another is the most powerful tool to create zeros and simplify the matrix structure.

Question

The values of $	heta$ lying between $	heta = 0$ and $	heta = \pi/2$ and satisfying the equation $\begin{vmatrix} 1 + \sin^2 	heta & \cos^2 	heta & 4 \sin 4	heta \ \sin^2 	heta & 1 + \cos^2 	heta & 4 \sin 4	heta \ \sin^2 	heta & \cos^2 	heta & 1 + 4 \sin 4	heta \end{vmatrix} = 0$ are

Options:

(A) $7\pi/24$

(B) $5\pi/24$

(C) $11\pi/24$

(D) $\pi/24$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating wall of numbers and trigonometric functions. This is the determinant we are facing:

$$
\begin{vmatrix} 1 + \sin^2 	heta & \cos^2 	heta & 4 \sin 4	heta \ \sin^2 	heta & 1 + \cos^2 	heta & 4 \sin 4	heta \ \sin^2 	heta & \cos^2 	heta & 1 + 4 \sin 4	heta \end{vmatrix} = 0
$$

It looks like a fortress, doesn't it? But every fortress has a weak point. In linear algebra, that weak point is often the structure of the rows or columns.

Instead of charging head-first into a direct expansion, which would lead to a chaotic mess of terms, let us be strategic.

### The Surgical Strike: Row Operations

Our first move is to simplify. We notice that the rows are remarkably similar.

If we subtract row two from row one ($R_1 	o R_1 - R_2$), the $\sin^2 	heta$ and $\cos^2 	heta$ terms begin to collapse. The first row becomes $[1, -1, 0]$.

This is the beauty of row operations—we are not changing the value of the determinant, but we are making it reveal its secrets. We repeat this for the second row ($R_2 	o R_2 - R_3$), and suddenly, the matrix is transformed into:

$$
\begin{vmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ \sin^2 	heta & \cos^2 	heta & 1 + 4 \sin 4	heta \end{vmatrix} = 0
$$

The complexity has vanished, replaced by a clean, manageable structure.

### The Elegant Expansion

Now that we have zeros, expanding along the first row is a breeze. We take the first element, $1$, and multiply it by the minor, then subtract the second element, $-1$, multiplied by its minor.

The calculation flows naturally:

$$
1[1(1 + 4 \sin 4	heta) - (-1)(\cos^2 	heta)] - (-1)[0(1 + 4 \sin 4	heta) - (-1)(\sin^2 	heta)] = 0
$$

Simplifying this, we get:

$$
(1 + 4 \sin 4	heta + \cos^2 	heta) + (\sin^2 	heta) = 0
$$

### The Final Bridge

Here is where the magic of trigonometry ties it all together. We see $\cos^2 	heta + \sin^2 	heta$.

We know this identity is the bedrock of trigonometry: $\sin^2 	heta + \cos^2 	heta = 1$. Substituting this in, the equation collapses into:

$$
1 + 4 \sin 4	heta + 1 = 0
$$

This simplifies to $2 + 4 \sin 4	heta = 0$. Solving for $\sin 4	heta$, we get:

$$
\sin 4	heta = -\frac{1}{2}
$$

### The Final Domain

We are almost at the finish line. We need to solve for $	heta$ where $0 < 	heta < \frac{\pi}{2}$. As we discussed, this means $0 < 4	heta < 2\pi$.

We are looking for angles where the sine is negative, which occurs in the third and fourth quadrants. Thus:

$$
4	heta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \quad 	ext{and} \quad 4	heta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}
$$

Dividing by $4$, we find our final values:

$$
	heta = \frac{7\pi}{24} \quad 	ext{and} \quad 	heta = \frac{11\pi}{24}
$$

Both values are within our range. You have successfully navigated the fortress and found the truth hidden within the numbers. Keep this mindset—look for the pattern, simplify, and then execute.

Animated Solution for Mathematics - Matrices and Determinants: The values of $θ$ lying between $θ = 0$ and $θ = π /2$ and satisfying the equation $1 + sin^{2} θ sin^{2} θ sin^{2} θ cos^{2} θ 1 + cos^{2} θ cos^{2} θ 4 sin 4 θ 4 sin 4 θ 1 + 4 sin 4 θ = 0$ are

Select Answer:

Visualized Solution

Original Determinant

Row Operation: $R_{1} \to R_{1} - R_{2}$

Row Operation: $R_{2} \to R_{2} - R_{3}$

Simplified Determinant

Expanding the Determinant

Applying $sin^{2} θ + cos^{2} θ = 1$

Solving for $sin 4 θ$

Finding $4 θ$ in $(0, 2 π)$

Final Values of $θ$

The Way Forward

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Surgical Strike

The Elegant Expansion

The Final Bridge

The Final Domain

Similar Questions

The solutions of the equation $1 + sin^{2} x cos^{2} x 4 sin 2 x sin^{2} x 1 + cos^{2} x 4 sin 2 x sin^{2} x cos^{2} x 1 + 4 sin 2 x = 0, (0 < x < π)$ are:

The values of $α$ , for which $11 2 α + 3 \frac{3}{2} \frac{1}{3} 3 α + 1 α + \frac{3}{2} α + \frac{1}{3} 0 = 0$ , lie in the interval

If $Δ_{1} = x - sin θ cos θ sin θ - x 1 cos θ 1 x$ and $Δ_{2} = x - sin 2 θ cos 2 θ sin 2 θ - x 1 cos 2 θ 1 x$ , $x \neq = 0$ ; then for all $θ \in (0, \frac{π}{2})$ :

Prove that for all values of $θ$ , $sin θ sin (θ + \frac{2 π}{3}) sin (θ - \frac{2 π}{3}) cos θ cos (θ + \frac{2 π}{3}) cos (θ - \frac{2 π}{3}) sin 2 θ sin (2 θ + \frac{4 π}{3}) sin (2 θ - \frac{4 π}{3}) = 0$ .

Which of the following values of $α$ satisfy the equation $(1 + α)^{2} (2 + α)^{2} (3 + α)^{2} (1 + 2 α)^{2} (2 + 2 α)^{2} (3 + 2 α)^{2} (1 + 3 α)^{2} (2 + 3 α)^{2} (3 + 3 α)^{2} = - 648 α$ ?

Let $m$ and $M$ be respectively the minimum and maximum values of $cos^{2} x 1 + cos^{2} x cos^{2} x 1 + sin^{2} x sin^{2} x sin^{2} x sin 2 x sin 2 x 1 + sin 2 x$ . Then the ordered pair $(m, M)$ is equal to :

Let $d \in R$ , and $A = - 2 15 4 + d (sin θ) + 2 (2 sin θ) - d (sin θ) - 2 d (- sin θ) + 2 + 2 d$ , $θ \in [0, 2 π]$ . If the minimum value of $det (A)$ is 8, then a value of d is :

If $A = 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1$ ; then for all $θ \in (3 π /4, 5 π /4)$ , $det (A)$ lies in the interval :

If the minimum and the maximum values of the function $f : [\frac{π}{4}, \frac{π}{2}] \to R$ , defined by $f (θ) = - sin^{2} θ - cos^{2} θ 12 - 1 - sin^{2} θ - 1 - cos^{2} θ 10 11 - 2$ are $m$ and $M$ respectively, then the ordered pair $(m, M)$ is equal to :

Let $f (x) = sin^{2} x 2 + sin^{2} x sin^{2} x - 2 + cos^{2} x cos^{2} x cos^{2} x cos 2 x cos 2 x 1 + cos 2 x, x \in [0, π]$ . Then the maximum value of $f (x)$ is equal to

Select Answer:

Visualized Solution

Original Determinant

Row Operation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​→R1​−R2​

Row Operation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​→R2​−R3​

Simplified Determinant

Expanding the Determinant

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ+cos2θ=1

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin4θ

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4θ in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,2π)

Final Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ