Question

The number of values of $\theta \in (0,\pi )$ for which the system of linear equations $x+3y+7z=0$
$-x+4y+7z=0$
$\left(\sin 3\theta \right)x+\left(\cos 2\theta \right)y+2z=0$ has a non-trivial solution, is :

• A. four
• B. three
• C. two
• D. one

Answer 1

The correct option is C two

For non-trivial solution
$\left|\begin{array}{ccc}1& 3& 7\\ -1& 4& 7\\ \sin 3\theta & \cos 2\theta & 2\end{array}\right|=0$

Let's expand along the row $1$
$\Rightarrow 1(8-7\cos 2\theta )-3(-2-7\sin 3\theta )+7(-\cos 2\theta -4\sin 3\theta )=0$
$\Rightarrow -2\cos 2\theta -\sin 3\theta +2=0$
$\Rightarrow 4{\sin }^3\theta +4{\sin }^2\theta -3\sin \theta =0$
$\Rightarrow \sin \theta (4{\sin }^2\theta +4\sin \theta -3)=0\Rightarrow \sin \theta (2\sin \theta +3)(2\sin \theta -1)=0$
$\Rightarrow \sin \theta =\frac{1}{2}\because \sin \theta \ne -\frac{3}{2}\text{and}\theta \in (0,\pi )$
$\Rightarrow \theta =\frac{\pi }{6}$ & $\frac{5\pi }{6}$ satisfy the equation
Hence, number of solutions $=2$