Consider the system of linear equations in x, y, z: (sin 3θ)x - y + z = 0, (cos 2θ)x + 4y + 3z = 0,

Q: Wait, why do we reject the negative root here?

Great catch! In our cubic equation, we found $\sin \theta = -\frac{3}{2}$. Remember the fundamental range of the sine function: for any real $\theta$, $-1 \le \sin \theta \le 1$. Since $-\frac{3}{2}$ is less than $-1$, it is mathematically impossible for $\sin \theta$ to take this value. We must reject it to avoid finding 'solutions' that don't exist in the real number system.

Q: Why can't we just use Cramer's Rule or Gaussian elimination?

Cramer's Rule is designed to find a unique solution when the determinant is non-zero. However, in this homogeneous system, if the determinant were non-zero, the only solution would be the trivial one ($x=0, y=0, z=0$). Since we are specifically hunting for *nontrivial* solutions, we are forced to set the determinant to zero. This is the only condition that allows for infinitely many solutions.

Q: How do I remember the general solution for $\sin \theta = 1/2$?

Don't memorize it blindly! The general solution for $\sin \theta = \sin \alpha$ is $\theta = n\pi + (-1)^n \alpha$. Here, $\sin \theta = 1/2$, so the principal value $\alpha$ is $\frac{\pi}{6}$. Plugging that into the formula gives you $\theta = n\pi + (-1)^n \frac{\pi}{6}$. It's all about understanding the periodicity of the sine wave!

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are going to peel back the layers of a classic JEE Advanced problem. It is a beautiful intersection of linear algebra and trigonometry.

We are presented with a system of three linear equations in $x$, $y$, and $z$:
$$(\sin 3	heta)x - y + z = 0$$
$$(\cos 2	heta)x + 4y + 3z = 0$$
$$2x + 7y + 7z = 0$$

Notice something crucial? The right-hand side of every single equation is zero. This is what we call a **homogeneous system**.

In the world of linear algebra, a homogeneous system $AX = 0$ always possesses the trivial solution where $x=y=z=0$. However, we are looking for **nontrivial solutions**—cases where the variables are not all zero.

For this to happen, the coefficient matrix $A$ must be singular. In plain English, the determinant of the coefficient matrix must be exactly zero: $|A| = 0$. This is the key that unlocks the entire problem.

### The Determinant Dance

Let us construct our determinant $\Delta$ using the coefficients of $x$, $y$, and $z$:

$$\Delta = \begin{vmatrix} \sin 3	heta & -1 & 1 \ \cos 2	heta & 4 & 3 \ 2 & 7 & 7 \end{vmatrix} = 0$$

Now, take a deep breath. Expanding a $3 	imes 3$ determinant can be messy if you rush, so let us be methodical. Expanding along the first row, we get:

$$\sin 3	heta (28 - 21) - (-1) (7\cos 2	heta - 6) + 1 (7\cos 2	heta - 8) = 0$$

Simplifying this, we find:

$$7\sin 3	heta + (7\cos 2	heta - 6) + (7\cos 2	heta - 8) = 0$$

Combining like terms, we arrive at:

$$7\sin 3	heta + 14\cos 2	heta - 14 = 0$$

See how the numbers align? Everything is a multiple of seven. Dividing by seven, we get the elegant equation:

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

This is the bridge between our algebra and our trigonometry.

### The Trigonometric Bridge

We have an equation with two different angles: $3	heta$ and $2	heta$. To solve this, we must unify them by expressing everything in terms of $\sin 	heta$.

We recall our trusty identities:
$$\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$$
$$\cos 2	heta = 1 - 2\sin^2 	heta$$

Substituting these into our equation, we get:

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

Expanding this, the constants $2$ and $-2$ cancel out beautifully, leaving us with:

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

This is a cubic equation in terms of $\sin 	heta$. Let us factor out $\sin 	heta$ to see what lies beneath:

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

Multiplying by $-1$ for clarity, we have:

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

### The Final Taming

We are almost there. We have two factors: $\sin 	heta = 0$ and the quadratic $4\sin^2 	heta + 4\sin 	heta - 3 = 0$.

Splitting the middle term of the quadratic, we get:

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

This gives us three potential paths: $\sin 	heta = 0$, $\sin 	heta = \frac{1}{2}$, and $\sin 	heta = -\frac{3}{2}$.

As we discussed, we must reject $\sin 	heta = -\frac{3}{2}$ because the sine function is bounded between $-1$ and $1$.

For $\sin 	heta = 0$, the general solution is $	heta = n\pi$. For $\sin 	heta = \frac{1}{2}$, the principal angle is $\frac{\pi}{6}$, leading to the general solution $	heta = n\pi + (-1)^n \frac{\pi}{6}$.

These are the values of $	heta$ that allow our system to breathe and exist in a nontrivial state. You have navigated the algebra, mastered the identities, and avoided the traps. That is the essence of JEE math—not just calculation, but insight.

Animated Solution for Mathematics - Matrices and Determinants: 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.

Visualized Solution

Analyze the System of Equations

Condition for Nontrivial Solutions

Construct the Determinant $Δ$

Expand the Determinant

Simplify the Equation

Apply Trigonometric Identities

Substitute and Expand

Factorize the Expression

Solve for $sin θ$

General Solutions and Conclusion

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

Analyzing the Setup

The Determinant Dance

The Trigonometric Bridge

The Final Taming

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