Question

The value of $\theta$ satisfying $\left|\begin{array}{ccc}1& 1& \sin 3\theta \\ -4& 3& \cos 2\theta \\ 7& -7& -2\end{array}\right|=0$ in $[0,2\pi ]$ is

• A. 5
• B. 5.00
• C. 5.000
• D. 5.0

Answer 1

Answer: (A), (B), (C), (D)

We have
$\left|\begin{array}{ccc}1& 1& \sin 3\theta \\ -4& 3& \cos 2\theta \\ 7& -7& -2\end{array}\right|=0$
Applying $C_1\to C_1+C_2,$ we get
$\left|\begin{array}{ccc}2& 1& \sin 3\theta \\ -1& 3& \cos 2\theta \\ 0& -7& -2\end{array}\right|=0$
Applying $R_1\to R_1+2R_2$
$\left|\begin{array}{ccc}0& 7& \sin 3\theta +2\cos 2\theta \\ -1& 3& \cos 2\theta \\ 0& -7& -2\end{array}\right|=0$
Expanding along $C_1,$ we get
$=1[-14+7(\sin 3\theta +2\cos 2\theta ]=0$
$\Rightarrow (3\sin \theta -4{\sin }^3\theta )+2(1-2{\sin }^2\theta )-2=0\Rightarrow 4{\sin }^3\theta +4{\sin }^2\theta -3\sin \theta =0\Rightarrow \sin \theta (4{\sin }^2\theta +4\sin \theta -3)=0\Rightarrow \sin \theta (4{\sin }^2\theta +6\sin \theta -2\sin \theta -3)=0\Rightarrow \sin \theta (2\sin \theta -1)(2\sin \theta +3)=0\Rightarrow \sin \theta =0\text{or}\sin \theta =\frac{1}{2}\text{or}\sin \theta =-\frac{3}{2}$
$\Rightarrow \sin \theta \ne \frac{-3}{2}$
$\therefore \sin \theta =0$ or $\frac{1}{2}$
$\therefore$ Possible values of $\theta =0,\pi ,2\pi ,\frac{\pi }{6},\frac{5\pi }{6}$