If the system of equations

x+(sqrt{2} sin | vedclass

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Question

$\left|\begin{array}{ccc}1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \ 1 & \sin \alpha & -\cos \alpha \ 1 & \cos \alpha &amp

Vedclass · Accepted Answer

$\frac{5 \pi}{24}$

Vedclass · Answer

$\left|\begin{array}{ccc}1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \alpha \ 1 & \sin \alpha & -\cos \alpha \ 1 & \cos \alpha & \sin \alpha\end{array}ight|=0$

$ \Rightarrow 1-\sqrt{2} \sin \alpha(\sin \alpha+\cos \alpha)+\sqrt{2} \cos \alpha(\cos \alpha-\sin \alpha)=0 $

$ \Rightarrow 1+\sqrt{2} \cos 2 \alpha-\sqrt{2} \sin 2 \alpha=0 $

$ \cos 2 \alpha-\sin 2 \alpha=-\frac{1}{\sqrt{2}} $

$ \cos \left(2 \alpha+\frac{\pi}{4}ight)=-\frac{1}{2} $

$ 2 \alpha+\frac{\pi}{4}=2 \mathrm{n} \pi \pm \frac{2 \pi}{3} $

$ \alpha+\frac{\pi}{8}=\mathrm{n} \pi \pm \frac{\pi}{3} $

$ \mathrm{n}=0 $

$ \mathrm{x}=\frac{\pi}{3}-\frac{\pi}{8}=\frac{5 \pi}{24}$

If the system of equations

$ x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 $

$ x+(\cos \alpha) y+(\sin \alpha) z=0 $

$ x+(\sin \alpha) y-(\cos \alpha) z=0$

has a non-trivial solution, then $\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :

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