The number of θ ∈ (0, 4π) for which the system of linear equations 3 (sin 3θ) x - y + z = 2, 3 (cos

Q: Wait, why do we only check $D=0$? What about $D_x, D_y, D_z$?

That is a brilliant observation! For a system to have 'no solution', the condition is indeed $D=0$ AND at least one of $D_x, D_y, D_z \neq 0$. In many JEE problems of this type, the values of $\theta$ that make $D=0$ automatically make at least one of the other determinants non-zero. If you were to check them, you would find they are non-zero for our solutions, confirming the 'no solution' case.

Q: Why did we reject $\sin \theta = -\frac{3}{2}$?

Remember the fundamental property of the sine function: for any real angle $\theta$, the value of $\sin \theta$ must lie within the closed interval $[-1, 1]$. Since $-\frac{3}{2} = -1.5$, which is less than $-1$, it is impossible for $\sin \theta$ to take this value. Always keep this constraint in mind when solving trigonometric equations!

Q: How do I handle the interval $(0, 4\pi)$ correctly?

The interval $(0, 4\pi)$ represents two full rotations around the unit circle. When you find a value like $\sin \theta = \frac{1}{2}$, you look for angles in the first and second quadrants. In the first rotation $(0, 2\pi)$, you get $\frac{\pi}{6}$ and $\frac{5\pi}{6}$. In the second rotation $(2\pi, 4\pi)$, you simply add $2\pi$ to these values, giving you $\frac{13\pi}{6}$ and $\frac{17\pi}{6}$. Always visualize the unit circle to ensure you don't miss any solutions!

Question

The number of $	heta \in (0, 4\pi)$ for which the system of linear equations $3 (\sin 3	heta) x - y + z = 2$, $3 (\cos 2	heta) x + 4y + 3z = 3$, $6x + 7y + 7z = 9$ has no solution is :

Options:

(A) 6

(B) 7

(C) 8

(D) 9

Sigma Percentile · Accepted Answer

### The Geometry of Impossibility: Unlocking the System

Imagine you are standing in a three-dimensional space, looking at three giant, flat planes. Each of your equations represents one of these planes.

Usually, these planes intersect at a single, unique point—a perfect, singular solution. But today, we are hunting for the moments when this harmony breaks down.

We are looking for the values of $	heta$ where the system has 'no solution'. This is the mathematical equivalent of the planes being parallel or forming a prism where they never all meet at once. To find these elusive values, we turn to the powerful tool of **Cramer's Rule**.

### The Determinant: Our First Gatekeeper

We begin by constructing the determinant $D$ from the coefficients of our variables $x, y,$ and $z$. The matrix is defined as:

$$D = \begin{vmatrix} 3 \sin 3	heta & -1 & 1 \ 3 \cos 2	heta & 4 & 3 \ 6 & 7 & 7 \end{vmatrix}$$

For the system to have no solution, the first requirement is that this determinant must vanish: $D = 0$.

Expanding this determinant along the first row requires careful attention to detail:

$$3 \sin 3	heta (28 - 21) + 1 (21 \cos 2	heta - 18) + 1 (21 \cos 2	heta - 24) = 0$$

Simplifying this, we get $21 \sin 3	heta + 42 \cos 2	heta - 42 = 0$. Dividing the entire equation by $21$ leaves us with a much cleaner expression:

$$\sin 3	heta + 2 \cos 2	heta - 2 = 0$$

This is our bridge between linear algebra and trigonometry.

### Unifying the Angles

Now, we face a classic JEE challenge: mixed angles. We have $3	heta$ and $2	heta$. To solve this, we must unify them.

We use the triple-angle identity $\sin 3	heta = 3 \sin 	heta - 4 \sin^3 	heta$ and the double-angle identity $\cos 2	heta = 1 - 2 \sin^2 	heta$. Substituting these into our equation, we get:

$$(3 \sin 	heta - 4 \sin^3 	heta) + 2(1 - 2 \sin^2 	heta) - 2 = 0$$

Watch the magic happen as the constants cancel out:

$$3 \sin 	heta - 4 \sin^3 	heta + 2 - 4 \sin^2 	heta - 2 = 0$$

We are left with $3 \sin 	heta - 4 \sin^3 	heta - 4 \sin^2 	heta = 0$. Multiplying by $-1$ to make the leading coefficient positive, we arrive at the cubic equation:

$$4 \sin^3 	heta + 4 \sin^2 	heta - 3 \sin 	heta = 0$$

### The Final Count

Solving this cubic is simpler than it looks. Factor out $\sin 	heta$ to get:

$$\sin 	heta (4 \sin^2 	heta + 4 \sin 	heta - 3) = 0$$

Further factorizing the quadratic gives us:

$$\sin 	heta (2 \sin 	heta - 1)(2 \sin 	heta + 3) = 0$$

This yields three potential values for $\sin 	heta$: $0$, $\frac{1}{2}$, and $-\frac{3}{2}$. As we discussed, $\sin 	heta = -\frac{3}{2}$ is impossible, so we discard it.

Now, we look at the interval $(0, 4\pi)$. For $\sin 	heta = 0$, we have $	heta = \pi, 2\pi, 3\pi$ ($3$ solutions).

For $\sin 	heta = \frac{1}{2}$, we have $	heta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$ ($4$ solutions).

Adding these together, we find a total of **$7$ values** for $	heta$. You have successfully navigated the complexity of the system and emerged with the answer.

Animated Solution for Mathematics - Matrices and Determinants: The number of $θ \in (0, 4 π)$ for which the system of linear equations $3 (sin 3 θ) x - y + z = 2$ , $3 (cos 2 θ) x + 4 y + 3 z = 3$ , $6 x + 7 y + 7 z = 9$ has no solution is :

Select Answer:

Visualized Solution

Condition for No Solution

Constructing Determinant $D$

Expanding the Determinant

Simplifying the Equation

Applying Trigonometric Identities

Substitution and Expansion

The Cubic Equation

Factorizing the Expression

Finding Possible Sine Values

Solving $sin θ = 0$ in $(0, 4 π)$

Solving $sin θ = \frac{1}{2}$ in $(0, 4 π)$

Final Count of Solutions

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

The Geometry of Impossibility

The Determinant

Unifying the Angles

The Final Count

Similar Questions

Consider the system of linear equations in $x, y, z$ : $(sin 3 θ) x - y + z = 0, (cos 2 θ) x + 4 y + 3 z = 0, 2 x + 7 y + 7 z = 0$ . Find the values of $θ$ for which this system has nontrivial solutions.

The number of real values $λ$ , such that the system of linear equations $2 x - 3 y + 5 z = 9$ , $x + 3 y - z = - 18$ , $3 x - y + (λ^{2} - ∣ λ ∣) z = 16$ has no solution, is :-

Let $θ \in (0, \frac{π}{2})$ . If the system of linear equations $(1 + cos^{2} θ) x + sin^{2} θ y + 4 sin 3 θ z = 0$ , $cos^{2} θ x + (1 + sin^{2} θ) y + 4 sin 3 θ z = 0$ , $cos^{2} θ x + sin^{2} θ y + (1 + 4 sin 3 θ) z = 0$ has a non-trivial solution, then the value of $θ$ is :

If the system of linear equations $2 x + 3 y - z = - 2$ , $x + y + z = 4$ , $x - y + ∣ λ ∣ z = 4 λ - 4$ where $λ \in R$ , has no solution, then

The sum of distinct values of $λ$ for which the system of equations $(λ - 1) x + (3 λ + 1) y + 2 λ z = 0 (λ - 1) x + (4 λ - 2) y + (λ + 3) z = 0 2 x + (3 λ + 1) y + 3 (λ - 1) z = 0$ has non-zero solutions, is

The system of equations $x + y + z = 6$ $x + 2 y + 5 z = 9,$ $x + 5 y + λ z = μ,$ has no solution if

If the system of equations $3 x + y + 4 z = 3$ , $2 x + α y - z = - 3$ , $x + 2 y + z = 4$ has no solution, then the value of $α$ is equal to :

Let $S$ be the set of all values of $θ \in [- π, π]$ for which the system of linear equations $x + y + 3 z = 0$ , $- x + (tan θ) y + 7 z = 0$ , $x + y + (tan θ) z = 0$ has non-trivial solution. Then $120 \sum_{θ \in S} θ$ is equal to

If $α + β + γ = 2 π$ , then the system of equations $x + (cos γ) y + (cos β) z = 0$ , $(cos γ) x + y + (cos α) z = 0$ , $(cos β) x + (cos α) y + z = 0$ has :

If the system of linear equations $3 x + y + β z = 3$ , $2 x + α y - z = - 3$ , $x + 2 y + z = 4$ has infinitely many solutions, then the value of $22 β - 9 α$ is :