Question

The system of linear equations $(\sin\theta)x+y-2z=0$, $2x-y+(\cos\theta)z = 0$ and $-3x+(\sec\theta)y+3z=0$, where $\theta \neq (2n+1)\frac{\pi}{2}$, has non-trivial solution for

• A. no value of $\theta$
• B. $\theta = n\pi + \frac{\pi}{4}, n \in \mathbb{Z}$
• C. $\theta = \text{Tan}^{-1}\left(\frac{3}{4}\right)$
• D. $\theta = \text{Tan}^{-1}\left(\frac{4}{3}\right)$

Hint:

When solving a trigonometric equation, if you arrive at an expression of the form $a\sin x + b\cos x = c$, first check if a solution is possible. A solution exists only if $|c| \le \sqrt{a^2+b^2}$. In this problem, we have $4\sin(2\theta) + 3\cos(2\theta) = -11$. Here $|-11| = 11$ and $\sqrt{4^2+3^2} = 5$. Since $11>5$, there are no real solutions for $\theta$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
A system of homogeneous linear equations, represented as \( AX = 0 \), has a non-trivial solution (a solution other than \( x = y = z = 0 \)) if and only if the determinant of the coefficient matrix \( A \) is zero. 
Step 2: Key Formula or Approach 
The system of equations is:
\[ (\sin\theta)x + y - 2z = 0 \] \[ 2x - y + (\cos\theta)z = 0 \] \[ -3x + (\sec\theta)y + 3z = 0 \] We need to find the values of \( \theta \) for which \( \det(A) = 0 \). 
Step 3: Detailed Explanation 
The coefficient matrix \( A \) is: \[ A = \begin{pmatrix} \sin\theta & 1 & -2 \\ 2 & -1 & \cos\theta \\ -3 & \sec\theta & 3 \end{pmatrix} \] We set the determinant to zero: \[ \det(A) = \sin\theta \begin{vmatrix} -1 & \cos\theta \\ \sec\theta & 3 \end{vmatrix} - 1 \begin{vmatrix} 2 & \cos\theta \\ -3 & 3 \end{vmatrix} + (-2) \begin{vmatrix} 2 & -1 \\ -3 & \sec\theta \end{vmatrix} = 0 \] Let's compute the smaller determinants. Note that \( \sec\theta \cdot \cos\theta = 1 \) since \( \theta \neq (2n+1)\frac{\pi}{2} \). \[ \sin\theta(-3 - (\cos\theta)(\sec\theta)) - 1(6 - (\cos\theta)(-3)) - 2(2\sec\theta - (-1)(-3)) = 0 \] \[ \sin\theta(-3 - 1) - 1(6 + 3\cos\theta) - 2(2\sec\theta - 3) = 0 \] \[ -4\sin\theta - 6 - 3\cos\theta - 4\sec\theta + 6 = 0 \] \[ -4\sin\theta - 3\cos\theta - 4\sec\theta = 0 \] Substitute \( \sec\theta = \frac{1}{\cos\theta} \): \[ -4\sin\theta - 3\cos\theta - \frac{4}{\cos\theta} = 0 \] Multiply the entire equation by \( \cos\theta \) (which is non-zero):
\[ -4\sin\theta\cos\theta - 3\cos^2\theta - 4 = 0 \] \[ 4\sin\theta\cos\theta + 3\cos^2\theta + 4 = 0 \] Using trigonometric identities \( 2\sin\theta\cos\theta = \sin(2\theta) \) and \( \cos^2\theta = \frac{1 + \cos(2\theta)}{2} \): \[ 2(2\sin\theta\cos\theta) + 3\left(\frac{1 + \cos(2\theta)}{2}\right) + 4 = 0 \] \[ 2\sin(2\theta) + \frac{3}{2} + \frac{3}{2}\cos(2\theta) + 4 = 0 \] Multiply by 2 to clear the fraction:
\[ 4\sin(2\theta) + 3 + 3\cos(2\theta) + 8 = 0 \] \[ 4\sin(2\theta) + 3\cos(2\theta) + 11 = 0 \] Now, consider the expression \( 4\sin(2\theta) + 3\cos(2\theta) \). The range of an expression of the form \( a\sin x + b\cos x \) is \( [-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}] \).
Here, \( a = 4 \) and \( b = 3 \). The range is \( [-\sqrt{4^2+3^2}, \sqrt{4^2+3^2}] = [-\sqrt{25}, \sqrt{25}] = [-5, 5] \).
So, the minimum value of \( 4\sin(2\theta) + 3\cos(2\theta) \) is -5 and the maximum value is 5.
Therefore, the range of the expression \( 4\sin(2\theta) + 3\cos(2\theta) + 11 \) is:
Minimum value: \( -5 + 11 = 6 \)
Maximum value: \( 5 + 11 = 16 \)
The expression is always between 6 and 16, so it can never be equal to 0.
Step 4: Final Answer 
Since \( \det(A) \) can never be zero, the system of equations only has the trivial solution (\( x = y = z = 0 \)) for any value of \( \theta \). There is no value of \( \theta \) for which a non-trivial solution exists.