Question

Let $\lambda$ and $\alpha$ be real
$\lambda x+\sin \alpha \cdot y+\cos \alpha \cdot z=0$,
$x+\cos \alpha \cdot y+\sin \alpha \cdot z=0$,
$-x+\sin \alpha \cdot y-\cos \alpha \cdot z=0$
has a non-trivial solution. If $\lambda =1$, then all the values of $\alpha$ is equals.

• A. $n\pi +\frac{\pi }{2}$
• B. $n\pi +\frac{\pi }{4},n\pi$
• C. $n\pi \pm \frac{\pi }{2}$
• D. $n\pi +\frac{\pi }{4},\frac{\pi }{4}$

Answer 1

The correct option is B $n\pi +\frac{\pi }{4},n\pi$

The given system has a non-trivial solution, if
$\left|\begin{array}{ccc}\lambda & \sin \alpha & \cos \alpha \\ 1& \cos \alpha & \sin \alpha \\ -1& \sin \alpha & -\cos \alpha \end{array}\right|=0$
By expanding the determinant along $\left(C_1\right)$, we get
$\lambda (-{\cos }^2\alpha -{\sin }^2\alpha )-1(-\sin \alpha \cos \alpha -\sin \alpha \cos \alpha )$
$-1({\sin }^2\alpha -{\cos }^2\alpha )=0$
$\lambda =\sin 2\alpha +\cos \alpha$
For $\lambda =1,\sin 2\alpha +\cos 2\alpha =1$
$\Rightarrow \frac{1}{\sqrt{2}}\sin 2\alpha +\frac{1}{\sqrt{2}}\cos 2\alpha =\frac{1}{\sqrt{2}}$
$\Rightarrow \cos (2\alpha -\frac{\pi }{4})=\cos (2n\pi \pm \frac{\pi }{4})$
$\Rightarrow 2\alpha =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{4},n$ being an integer.
$\Rightarrow \alpha =n\pi +\frac{\pi }{4},n\pi$.