Question

Let $\theta \in (0,\frac{\pi }{2}).$ If the system of linear equations,
$\begin{array}{c}(1+{\cos }^2\theta )x+{\sin }^2\theta y+4\sin 3\theta z=0\\ {\cos }^2\theta x+(1+{\sin }^2\theta )y+4\sin 3\theta z=0\\ {\cos }^2\theta x+{\sin }^2\theta y+(1+4\sin 3\theta )z=0\end{array}$
has a non-trivial solution, then the value of $\theta$ is

• A. $\frac{7\pi }{18}$
• B. $\frac{\pi }{18}$
• C. $\frac{4\pi }{9}$
• D. $\frac{5\pi }{18}$

Answer 1

The correct option is A $\frac{7\pi }{18}$

$\Delta =\left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\sin 3\theta \\ {\cos }^2\theta & 1+{\sin }^2\theta & 4\sin 3\theta \\ {\cos }^2\theta & {\sin }^2\theta & 1+4\sin 3\theta \end{array}\right|C_1\to C_1+C_2+C_3\Rightarrow \Delta =\left|\begin{array}{ccc}2+4\sin 3\theta & {\sin }^2\theta & 4\sin 3\theta \\ 2+4\sin 3\theta & 1+{\sin }^2\theta & 4\sin 3\theta \\ 2+4\sin 3\theta & {\sin }^2\theta & 1+4\sin 3\theta \end{array}\right|\Rightarrow \Delta =(2+4\sin 3\theta )\left|\begin{array}{ccc}1& {\sin }^2\theta & 4\sin 3\theta \\ 1& 1+{\sin }^2\theta & 4\sin 3\theta \\ 1& {\sin }^2\theta & 1+4\sin 3\theta \end{array}\right|R_2\to R_2-R_1\text{and}{R}_3\to R_3-R_1\Rightarrow \Delta =\left|\begin{array}{ccc}1& {\sin }^2\theta & 4\sin 3\theta \\ 0& 1& 0\\ 0& 0& 1\end{array}\right|\times (2+4\sin 3\theta )\Rightarrow \Delta =(2+4\sin 3\theta )$
For non-trivial solution
$\Delta =0$
$\Rightarrow \sin 3\theta =\frac{-1}{2}$
$\Rightarrow \theta =\frac{7\pi }{18}$