Question

The system of equations
$x-y\cos \theta +z\cos 2\theta =0-x\cos \theta +y-z\cos \theta =0x\cos 2\theta -y\cos \theta +z=0$
has non-trivial solution for $\theta$ equals

• A. $n\pi$ only, $n\in Z$
• B. $n\pi +\frac{\pi }{4}$ only, $n\in Z$
• C. $(2n-1)\frac{\pi }{2}$ only, $n\in Z$
• D. all real values of $\theta$

Answer 1

The correct option is D all real values of $\theta$

For non-trivial solutions,
$\Delta =\left|\begin{array}{ccc}1& -\cos \theta & \cos 2\theta \\ -\cos \theta & 1& -\cos \theta \\ \cos 2\theta & -\cos \theta & 1\end{array}\right|=0$

Using $C_1\to C_1-C_3,$
$\Delta =\left|\begin{array}{ccc}2{\sin }^2\theta & -\cos \theta & \cos 2\theta \\ 0& 1& -\cos \theta \\ -2{\sin }^2\theta & -\cos \theta & 1\end{array}\right|=0$

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

$\Rightarrow {\sin }^2\theta =0,$
or $1[1-{\cos }^2\theta ]-1[{\cos }^2\theta -\cos 2\theta ]=0$
$\Rightarrow {\sin }^2\theta -[{\cos }^2\theta -({\cos }^2\theta -{\sin }^2\theta )]=0$
$\Rightarrow {\sin }^2\theta -{\sin }^2\theta =0$ which is always true.
Hence, $\Delta =0\forall \theta \in R$