Question

Find the value of $\theta$ satisfying $\left|\begin{array}{ccc}1& 1& sin3\theta \\ -4& 3& cos2\theta \\ 7& -7& -2\end{array}\right|$ = 0

Answer 1

We have, $\left|\begin{array}{ccc}1& 1& sin3\theta \\ -4& 3& cos2\theta \\ 7& -7& -2\end{array}\right|$
$\Rightarrow \left|\begin{array}{ccc}0& 1& sin3\theta \\ -7& 3& cos2\theta \\ 14& -7& -2\end{array}\right|=0$ $[\because C_1\to C_1-C_2]$
$\Rightarrow 7\left|\begin{array}{ccc}0& 1& sin3\theta \\ -1& 3& cos2\theta \\ 2& -7& -2\end{array}\right|=0$ [taking 7 common from $C_1$]
$\Rightarrow 7[0-1(2-2cos2\theta +sin3\theta (7-6)]=0$ [expanding along $R_1$]
$\Rightarrow 7[-2(1-cos2\theta )+sin3\theta ]=0\Rightarrow -14+14cos2\theta +7sin3\theta =0\Rightarrow 14cos2\theta +7sin3\theta =14\Rightarrow 14(1-2si{n}^2\theta )+7(3sin\theta -4si{n}^3\theta )=14\Rightarrow -28si{n}^2\theta +14+21sintheta-28si{n}^3\theta =14\Rightarrow -28si{n}^3\theta +28si{n}^2\theta -21sin\theta =0\Rightarrow 4si{n}^3\theta +4si{n}^2\theta -3sin\theta =0\Rightarrow sin\theta (4si{n}^2\theta +4sin\theta -3)=0\Rightarrow \text{Either}sin\theta =0\Rightarrow \theta =n\pi \text{or}4si{n}^2\theta +4sin\theta -3=0\therefore sin\theta =\frac{-4\pm \sqrt{16+48}}{8}=\frac{-4\pm \sqrt{64}}{8}=\frac{-4\pm 8}{8}=\frac{4}{8},\frac{-12}{8}sin\theta =\frac{1}{2},\frac{-3}{2}$
If $sin\theta =\frac{1}{2}=sin\frac{\pi }{6}$, then
$\theta =n\pi +(-1{)}^n\frac{\pi }{6}$
Hence, $sin\theta =\frac{-3}{2}$ not possible because $-1\le sin\theta \le 1$