Question

The value of ` $\theta$' for which the system of linear equation in $x,y,z$ given as $\left(\sin 3\theta \right)x-y+z=0$

$\left(\cos 2\theta \right)x+4y+3z=0,2x+7y+7z=0$ has a non trivial solution

• A. $\frac{n\pi }{2}$
• B. $\frac{n\pi }{3}$
• C. $n\pi +(-1{)}^n\frac{\pi }{6}$
• D. $\frac{\pi }{12}$

Answer 1

The correct option is C $n\pi +(-1{)}^n\frac{\pi }{6}$

Solutions are non-trivial
$\Rightarrow \left|\begin{array}{ccc}\sin 3\theta & -1& 1\\ \cos 2\theta & 4& 3\\ 2& 7& 7\end{array}\right|=0$
$\sin 3\theta \left(7\right)+1(7\cos 2\theta -6)+1(7\cos 2\theta -8)=0$
$7\cos 2\theta +7\sin 3\theta +7\cos 2\theta -14=0$
$7(2\cos 2\theta +7\sin 3\theta )=14$
$2(1-2{\sin }^2\theta )+3\sin \theta -4{\sin }^3\theta =2$
$3\sin \theta -4{\sin }^3\theta -4{\sin }^2\theta =0$
$\sin \theta (3-4{\sin }^2\theta -4\sin \theta )=0$
$\sin \theta =0and4{\sin }^2\theta +4\sin \theta -3=0$
$\sin \theta =\frac{-4\pm \sqrt{16+48}}{8}$
$=\frac{-4+8}{8}$
$\therefore \sin \theta =\frac{1}{2}$

$\therefore \theta =n\pi +(-1{)}^n\frac{\pi }{6}$ or $\theta =n\pi$