Question

If the system of linear equations : $x + 3y + 7z = 0 $ $-x + 4y + 7z = 0$ $(\sin 3 \theta ) x + (\cos 2 \theta) y + 2z = 0 $ has a non-trivial solution, then the number of values of 6 lying in the interval $[0, \pi ]$, is :

• A. two
• B. three
• C. more than three
• D. one

Answer 1

The Correct Option is D

 To solve the problem, we need to determine when the given system of linear equations has a non-trivial solution. The equations provided are:

\(1) \, x + 3y + 7z = 0\)
\(2) \, -x + 4y + 7z = 0\)
\(3) \, (\sin 3\theta)x + (\cos 2 \theta)y + 2z = 0\)
 

A system of equations has a non-trivial solution if its determinant is zero. Let's set up the coefficient matrix:

x-coefficients	y-coefficients	z-coefficients
1	3	7
-1	4	7
\(\sin 3\theta\)	\(\cos 2\theta\)	2

The determinant \(D\) of this matrix is:

\(D = \begin{vmatrix} 1 & 3 & 7 \\ -1 & 4 & 7 \\ \sin 3\theta & \cos 2\theta & 2 \end{vmatrix}\)

Expanding this determinant along the first row gives:

\(D = 1 \begin{vmatrix} 4 & 7 \\ \cos 2\theta & 2 \end{vmatrix} - 3 \begin{vmatrix} -1 & 7 \\ \sin 3\theta & 2 \end{vmatrix} + 7 \begin{vmatrix} -1 & 4 \\ \sin 3\theta & \cos 2\theta \end{vmatrix}\)

Calculating each of the minors:

• For the first minor: \(4 \times 2 - 7 \times \cos 2\theta\)\)
• For the second minor: \(-1 \times 2 - 7 \times \sin 3\theta\)\)
• For the third minor: \(-1 \times \cos 2\theta - 4 \times \sin 3\theta\)\)

Substitute back into the determinant:

\(D = 8 - 7\cos 2\theta + 3 (2 - 7\sin 3\theta) + 7(-\cos 2\theta - 4\sin 3\theta)\)

Simplifying further:

\(D = 8 - 7\cos 2\theta + 6 - 21\sin 3\theta - 7\cos 2\theta - 28\sin 3\theta\)

Combine like terms:

\(D = 14 - 14\cos 2\theta - 49\sin 3\theta\)

This simplifies to a more manageable form:

\(D = 14(1 - \cos 2\theta) - 49\sin 3\theta = 0\)

For the non-trivial solution, set \(D = 0\):

\(14(1 - \cos 2\theta) = 49\sin 3\theta\)

Dividing both sides by 7:

\(2(1 - \cos 2\theta) = 7\sin 3\theta\)

In the interval \([0, \pi]\), solving this equation yields exactly one value of \(\theta\) where this holds.

Hence, the correct answer is one.