Q46 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics
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
- A$\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}+\frac{\pi}{8}$ ( $n$ is an integer)
- B$\frac{n \pi}{2}+(-1)^n \frac{\pi}{8}$ ( $n$ is an integer)
- Correct answerC$\frac{n \pi}{2}+(-1)^n \frac{\pi}{8}-\frac{\pi}{8}$ ( $n$ is an integer)
- D$\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}-\frac{\pi}{8}$ ( $n$ is an integer)
Solution
$$
\begin{aligned}
x+\sin \alpha y+\cos \alpha z & =0, \quad x+\cos \alpha y+\sin \alpha z=0 \\
-x+\sin \alpha y-\cos \alpha z & =0
\end{aligned}
$$
$$
\begin{aligned}
& \text { Non-trivial sol, so }\left|\begin{array}{ccc}
1 & \sin \alpha & \cos \alpha \\
1 & \cos \alpha & \sin \alpha \\
-1 & \sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& \Rightarrow 1\left(-\sin ^2 \alpha-\cos ^2 \alpha\right)-\sin \alpha(-\cos \alpha+\sin \alpha) \\
& +\cos \alpha(\sin \alpha+\cos \alpha) \mid=0 \\
& -1-\sin \alpha(-\cos \alpha+\sin \alpha]+\cos \alpha[\sin \alpha+\cos \alpha]=0 \\
& -1+\sin \alpha \cos \alpha-\sin ^2 \alpha+\sin \alpha \cos \alpha+\cos ^2 \alpha=0 \\
& -1+\sin 2 \alpha+\cos 2 \alpha=0 \\
& \sin 2 \alpha+\cos 2 \alpha=1 \\
& \Rightarrow \quad \frac{1}{\sqrt{2}} \sin 2 \alpha+\frac{1}{\sqrt{2}} \cos 2 \alpha=\frac{1}{\sqrt{2}} \\
& \Rightarrow \quad \sin \left(2 \alpha+\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \\
&
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow \quad 2 \alpha+\frac{\pi}{4} & =n \pi+(-1)^n \frac{\pi}{4} \\
2 \alpha & =n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4} \\
\Rightarrow \quad \alpha & =\frac{n \pi}{2}+(-1)^n \frac{\pi}{8}-\frac{\pi}{8}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.