Answer:
Option C
Explanation:
We have,
$(4 \sin \theta )x-3y+z=0$
$ x-(6 \cos 2 \theta)y+z=0$
3x-12y+4z=0
system of equation has a non trival solution
$\therefore$ $\begin{bmatrix}4 \sin \theta &-3&1 \\1 & -6 \cos 2\theta&1\\3&-12&4 \end{bmatrix}=0$
$R_{1}\rightarrow 4R_{1}-R_{3}$
$\begin{bmatrix}16 \sin \theta-3 &0&0 \\3 & -6 \cos 2\theta&1\\3&-12&4 \end{bmatrix}=0$
= $(16 \sin \theta -3)(-24 \cos 2 \theta+12)=0$
=$(16 \sin \theta-3)(1-2 \cos \theta)=0$
$\Rightarrow$ $16 \sin \theta =1 or 2 \cos \theta$=1
$\theta = \sin^{1}\frac{3}{16}$ or $\theta$ = $\frac{\pi}{6}$
$\therefore$ $\sin^{-1} (\frac{3}{16})$
