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(C) For the system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix

Vedclass · Accepted Answer

$\frac{n \pi}{2}+(-1)^n \frac{\pi}{8}-\frac{\pi}{8}$ ($n$ is an integer)

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(C) For the system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix} 1 & \sin \alpha & \cos \alpha \ 1 & \cos \alpha & \sin \alpha \ -1 & \sin \alpha & -\cos \alpha \end{vmatrix} = 0$Expanding along the first row:$1(-\cos^2 \alpha - \sin^2 \alpha) - \sin \alpha(-\cos \alpha + \sin \alpha) + \cos \alpha(\sin \alpha + \cos \alpha) = 0$$-1 - \sin \alpha(-\cos \alpha + \sin \alpha) + \cos \alpha(\sin \alpha + \cos \alpha) = 0$$-1 + \sin \alpha \cos \alpha - \sin^2 \alpha + \sin \alpha \cos \alpha + \cos^2 \alpha = 0$$-1 + 2\sin \alpha \cos \alpha + (\cos^2 \alpha - \sin^2 \alpha) = 0$$-1 + \sin 2\alpha + \cos 2\alpha = 0$$\sin 2\alpha + \cos 2\alpha = 1$Dividing by $\sqrt{2}$:$\frac{1}{\sqrt{2}} \sin 2\alpha + \frac{1}{\sqrt{2}} \cos 2\alpha = \frac{1}{\sqrt{2}}$$\sin(2\alpha + \frac{\pi}{4}) = \sin \frac{\pi}{4}$General solution for $\sin 	heta = \sin \beta$ is $	heta = n\pi + (-1)^n \beta$.$2\alpha + \frac{\pi}{4} = n\pi + (-1)^n \frac{\pi}{4}$$2\alpha = n\pi + (-1)^n \frac{\pi}{4} - \frac{\pi}{4}$$\alpha = \frac{n\pi}{2} + (-1)^n \frac{\pi}{8} - \frac{\pi}{8}$

The set of real values of $\alpha$ for which the system of linear equations
$\begin{aligned}
& x+(\sin \alpha) y+(\cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& -x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$
has a non-trivial solution is

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The set of real values of $\alpha$ for which the system of linear equations$\begin{aligned}& x+(\sin \alpha) y+(\cos \alpha) z=0 \\& x+(\cos \alpha) y+(\sin \alpha) z=0 \\& -x+(\sin \alpha) y-(\cos \alpha) z=0\end{aligned}$has a non-trivial solution is