1112sinαsinαcosα2cosα−cosαsinα=0⇒1−2sinα(sinα+cosα)+2cosα(cosα−sinα)=0⇒1+2cos2α−2sin2α=0cos2α−sin2α=−21cos(2α+4π)=−212α+4π=2nπ±32πα+8π=nπ±3πn=0x=3π−8π=245π
If the system of equations $\begin{aligned}
& x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \
& x+(\cos \alpha) y+(\sin \alpha) z=0 \
& x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}hasanon−trivialsolution,then\alpha \in\left(0, \frac{\pi}{2}\right)$ is equal to :
Held on 4 Apr 2024 · Verified 6 Jul 2026.
2411π
245π
247π
43π
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