Q46 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics
A value of $\theta$ for which the following system of equations has a non-trivial solution is
$$
(4 \sin \theta) x-3 y-z=0, x-(6 \cos 2 \theta) y+z=0 \text {, }
$$
$$
3 x-12 y+4 z=0
$$
- A$\tan ^{-1}\left(\frac{1}{2}\right)$
- B$\frac{\pi}{4}$
- Correct answerC$\sin ^{-1}\left(\frac{3}{16}\right)$
- D$\frac{\pi}{12}$
Solution
We have,
$$
\begin{aligned}
& (4 \sin \theta) x-3 y+z=0 \\
& x-(6 \cos 2 \theta) y+z=0 \\
& 3 x-12 y+4 z=0
\end{aligned}
$$
System of equation has a non trivial solution
$$
\begin{aligned}
& \therefore \quad\left|\begin{array}{ccc}
4 \sin \theta & -3 & 1 \\
1 & -6 \cos 2 \theta & 1 \\
3 & -12 & 4
\end{array}\right|=0 \\
& R_1 \rightarrow 4 R_1-R_3 \\
& \left|\begin{array}{ccc}
16 \sin \theta-3 & 0 & 0 \\
3 & -6 \cos 2 \theta & 1 \\
3 & -12 & 4
\end{array}\right|=0 \\
& =(16 \sin \theta-3)(-24 \cos 2 \theta+12)=0 \\
& =(16 \sin \theta-3)(1-2 \cos \theta)=0 \\
& \Rightarrow 16 \sin \theta=1 \text { or } 2 \cos 2 \theta=1 \\
& \theta=\sin ^{-1} \frac{3}{16} \text { or } \theta=\frac{\pi}{6} \\
& \therefore \quad \theta=\sin ^{-1} \frac{3}{16} \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'C'.