Solve this
Question:

Let $\theta \in\left(0, \frac{\pi}{2}\right) .$ If the system of linear equations

$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$

$\cos ^{2} \theta x+\left(1+\sin ^{2} \theta\right) y+4 \sin 3 \theta z=0$

$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$

has a non-trivial solution, then the value of $\theta$ is :

1. $\frac{4 \pi}{9}$

2. $\frac{7 \pi}{18}$

3. $\frac{\pi}{18}$

4. $\frac{5 \pi}{18}$

Correct Option: , 2

Solution:

Case-I

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

$\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}$

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

$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}, \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}$

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

or $4 \sin 3 \theta=-2$

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

$\theta=\frac{7 \pi}{18}$