Mathematics - Determinant Question with Solution | TestHub

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

${\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{R}}_2$

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

$C_2\to C_2+C_1$

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

${\mathrm{C}}_2\to {\mathrm{C}}_2+{\mathrm{C}}_3$

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

Expanding along $R_3$

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

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

$\Rightarrow \sin 3\theta =\frac{-1}{2}$

$\Rightarrow 3\theta =\frac{7\pi }{6}$

$\Rightarrow \theta =\frac{7\pi }{18}$