Question

A value of $\theta \in (0,\frac{\pi }{3})$, for which $\left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\cos 6\theta \\ {\cos }^2\theta & 1+{\sin }^2\theta & 4\cos 6\theta \\ {\cos }^2\theta & {\sin }^2\theta & 1+4\cos 6\theta \end{array}\right|=0\text{is}:$

• A. $\frac{\pi }{18}$
• B. $\frac{\pi }{9}$
• C. $\frac{7\pi }{24}$
• D. $\frac{7\pi }{36}$

Answer 1

The correct option is B $\frac{\pi }{9}$

$\left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\cos 6\theta \\ {\cos }^2\theta & 1+{\sin }^2\theta & 4\cos 6\theta \\ {\cos }^2\theta & {\sin }^2\theta & 1+4\cos 6\theta \end{array}\right|=0$
Apply - $R_2\to R_2-R_1,R_3\to R_3-R_1$

$\Rightarrow \left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\cos 6\theta \\ -1& 1& 0\\ -1& 0& 1\end{array}\right|=0$

Apply - $C_1\to C_1+C_2$
$\Rightarrow \left|\begin{array}{ccc}2& {\sin }^2\theta & 4\cos 6\theta \\ 0& 1& 0\\ -1& 0& 1\end{array}\right|=0$
Expanding along $C_1$
$\Rightarrow 2(1-0)-1(-4\cos 6\theta )=0\Rightarrow \cos 6\theta =-\frac{1}{2}$
$\theta \in (0,\frac{\pi }{3})\Rightarrow 6\theta \in (0,2\pi )$
So,
$6\theta =\frac{2\pi }{3},\frac{4\pi }{3}\Rightarrow \theta =\frac{\pi }{9},\frac{2\pi }{9}$