Question

Find the values of $\theta$ such that the system of equations, $\left(\sin 3\theta \right)x-y+z=0,\left(\cos 2\theta \right)x+4y+3z=0,2x+7y+7z=0$ has non trivial solution.

• A. $n\pi ,n\in 1$
• B. $n\pi +{\left(-1\right)}^n\frac{\pi }{3},n\in 1$
• C. $n\pi +{\left(-1\right)}^n\frac{\pi }{6},n\in 1$
• D. Non of these

Answer 1

The correct option is B $n\pi +{\left(-1\right)}^n\frac{\pi }{3},n\in 1$

Since the given system has non-trivial solution.

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

$\Rightarrow \sin 3\theta (28-21)+1(7\cos 2\theta -6)+1(7\cos 2\theta -8)$

$\Rightarrow 7\sin 3\theta +14\cos 2\theta -14=0$

$\Rightarrow \sin 3\theta -2\cos 2\theta -2=0$

$\Rightarrow 3\sin \theta -4{\sin }^3\theta +2(1-2{\sin }^2\theta )-2=0$

$\Rightarrow \sin \theta (4{\sin }^2\theta +4\sin \theta -3)=0$

$\Rightarrow \sin \theta =0\Rightarrow \theta =n\pi$ ........$\left(1\right)$

or $4{\sin }^2\theta +4\sin \theta -3=0$

$\Rightarrow (2\sin \theta -1)(2\sin \theta +3)=0$

$\Rightarrow \sin \theta =\frac{1}{2}\because \sin \theta =-\frac{3}{2}$ is not possible.

$\therefore \theta =n\pi +{(-1)}^n\frac{\pi }{6}$ ......$\left(2\right)$

$\therefore$From equations $\left(1\right)$ and $\left(2\right)$, we get

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