Question

Consider the system of linear equation in x, y, z
$\left(sin3\theta \right)x-y+z=0$
$\left(cos2\theta \right)x+4y+3z=0$
$2x+7y+7z=0$
Find the values of $\theta$ for which this system has non trivial solution

• A. $\theta =n\pi$ or $\theta =m\pi +(-1{)}^m\pi /6,n,m\epsilon I$
• B. $\theta =n\pi$ or $\theta =m\pi +(-1{)}^m\pi /3,n,m\epsilon I$
• C. $\theta =n\pi$ or $\theta =m\pi +(-1{)}^m\pi /2,n,m\epsilon I$
• D. $\theta =n\pi$ or $\theta =m\pi +(-1{)}^m\pi /4,n,m\epsilon I$

Answer 1

The correct option is A $\theta =n\pi$ or $\theta =m\pi +(-1{)}^m\pi /6,n,m\epsilon I$

Since, the given system has non-trivial solution
$\therefore \left|\begin{array}{ccc}\sin 3\theta & -1& 1\\ \cos 2\theta & 4& 3\\ 2& 7& 7\end{array}\right|=0$
$\Rightarrow \sin 3\theta (28-21)+1(7\cos 2\theta -6)+1(7\cos 2\theta -8)+7\sin 3\theta +14\cos 2\theta -14=0\Rightarrow \sin 3\theta +2\cos 2\theta -2=0\Rightarrow 3\sin \theta -4{\sin }^3\theta +2(1-2{\sin }^2\theta )-2=0\Rightarrow \sin \theta (4{\sin }^2\theta +4\sin \theta -3)=0$
$\Rightarrow \sin \theta =0$ or $4{\sin }^2\theta +4\sin \theta -3=0$
$\Rightarrow \theta =n\pi$ or $(2\sin \theta -1)(2\sin \theta +3)=0$
$\Rightarrow \theta =n\pi$ or $\sin \theta =\frac{1}{2}$
$\Rightarrow \theta =n\pi$ or $\theta =n\pi +{(-1)}^n\left(\frac{\pi }{6}\right)$