Question

The value of $\theta$ lying between $\theta =0$ and $\frac{\pi }{2}$ and satisfying the equation.
$\left|\begin{array}{ccc}1+{\sin }^2\theta & {\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & 1+{\cos }^2\theta & 4\sin 4\theta \\ {\sin }^2\theta & {\cos }^2\theta & 1+4\sin 4\theta \end{array}\right|=0$ are

• A. $\frac{7\pi }{24}$
• B. $\frac{5\pi }{24}$
• C. $\frac{11\pi }{24}$
• D. $\frac{\pi }{24}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{7\pi }{24}$
C $\frac{11\pi }{24}$

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

Applying $R_3\to R_3-R_1$ and $R_2\to R_2-R_1$, we get

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

Applying $C_1\to C_1+C_2$

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

$\Rightarrow 2+4\sin 4\theta =0\Rightarrow \sin 4\theta =-\frac{1}{2}$

$\Rightarrow 4\theta =n\pi +{(-1)}^n(-\frac{\pi }{6})$

$\Rightarrow \theta =n\frac{\pi }{4}+{(-1)}^{n+1}\left(\frac{\pi }{24}\right)$

So, $\theta =\frac{7\pi }{24},\frac{11\pi }{24}$ are the two values lying between $0$ and $\frac{\pi }{2}$