Question

If $-\pi <\theta <\pi$, find the number of solutions of
$\Delta =\left|\begin{array}{ccc}\cos \theta & \sin \theta & \cos \theta \\ -\sin \theta & \cos \theta & \sin \theta \\ -\cos \theta & -\sin \theta & \cos \theta \end{array}\right|=0$

Answer 1

Given $\Delta =\left|\begin{array}{ccc}\cos \theta & \sin \theta & \cos \theta \\ -\sin \theta & \cos \theta & \sin \theta \\ -\cos \theta & -\sin \theta & \cos \theta \end{array}\right|=0$
$\Rightarrow \cos \theta \left(1\right)-\sin \theta \left(0\right)+\cos \theta \left(1\right)=0$

$\Rightarrow 2\cos \theta =0$

Since, $-\pi <\theta <\pi$

So, $\theta =-\frac{\pi }{2},\frac{\pi }{2}$

So, number of solutions $=2$