Question

Consider the system of equations in x, y, z as
$x\, sin \,3\theta - y + z = 0, \,x \,cos\, 2\theta + 4y + 3z = 0$,
$2x + 7y + 7z = 0.$ If this system has a non-trivial solution, then for integer n, values of $\theta$ are given by

• A. $\pi\left(n+\frac{\left(-1\right)^{n}}{3}\right)$
• B. $\pi\left(n+\frac{\left(-1\right)^{n}}{4}\right)$
• C. $\pi\left(n+\frac{\left(-1\right)^{n}}{6}\right)$
• D. $\frac{n\pi}{2}$

Answer 1

Answer: (C)

Given system of equations
$x \,sin\, 3\theta -y+ z =0$, $x \,cos\, 2\theta +4y +3z = 0$, $2x+7y +7z =0 \to$ are homogeneous system of linear equation
Since system has non-trivial solution
$\therefore \begin{vmatrix}sin\,3\theta&-1&1\\ cos\,2\theta &4&3\\ 2&7&7\end{vmatrix} = 0$
$\Rightarrow \quad sin \,3\theta \left[28 - 21\right] + 1 \left[7\, cos \,2\theta - 6\right] + \left[7 \,cos \,2\theta - 8\right] = 0$
$\Rightarrow \quad 3\, sin\,\theta - 4 \,sin^{3}\, \theta + 2\left(1 - 2 \,sin^{2} \,\theta \right) - 2 = 0$
$\Rightarrow \quad sin \,\theta \left(4 \,sin^{2} \,\theta + 4 \,sin \,\theta - 3\right) = 0$
Either $sin\, \theta = 0$ or $4\, sin2 \theta + 6\, sin \,\theta - 2\, sin\, \theta - 3 = 0$
$\Rightarrow \quad \left(2 \,sin\, \theta - 1\right) \left(2 \,sin \,\theta + 3\right) = 0$
$\therefore \quad sin \,\theta = \frac{1}{2}, sin \,\theta \ne -\frac{3}{2}\quad\left[\because sin \,\theta > -1\right]$
$\therefore \quad \,\theta = n\pi$ or $\,\theta = n\pi + \left(-1\right)^{n} \frac{\pi}{6}$
$\Rightarrow \quad\theta = \pi\left[n+\frac{\left(-1\right)^{n}}{6}\right].$