Question

The number of values of $\theta \in (0,\pi )$ for which the system of linear equations
$x+3y+7z=0$
$x+4y+7z=0$
$\left(\sin 3\theta \right)x+\left(\cos 2\theta \right)y+2z=0$
has a non-trivial solution, is :

• A. One
• B. Three
• C. Four
• D. Two

Answer 1

Answer: (D)

Two

$\left|\begin{array}{ccc}1& 3& 7\\ -1& 4& 7\\ \sin 3\theta & \cos 2\theta & 2\end{array}\right|=0$

$(8-7\cos 2\theta )-3(-2-7\sin 3\theta )+7(-\cos 2\theta -4\sin 3\theta )=0$

$14-7\cos 2\theta +21\sin 3\theta -7\cos 2\theta -28\sin 3\theta =0$

$14-7\sin 3\theta -14\cos 2\theta =0$

$14-7(3\sin \theta -4{\sin }^3\theta )-14(1-2{\sin }^2\theta )=0$

$-21\sin \theta +28{\sin }^3\theta +28{\sin }^2\theta =0$

$7\sin \theta [-3+4{\sin }^2\theta +4\sin \theta ]=0$

$\sin \theta$,

$4{\sin }^2\theta +6\sin \theta -2\sin \theta -3=0$

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

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

$\sin \theta =\frac{-3}{2};\sin \theta =\frac{1}{2}$

Hence, $2$ solution in $(0,\pi )$