Question

The value of $\theta$ lying between 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. $\frac{7\pi }{24}$ or $\frac{11\pi }{24}$
• B. $\frac{7\pi }{24}$ or $\frac{5\pi }{24}$
• C. $\frac{5\pi }{24}$ or $\frac{\pi }{24}$
• D. None of these.

Answer 1

The correct option is A $\frac{7\pi }{24}$ or $\frac{11\pi }{24}$

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

$\Rightarrow R_1\to R_1-R_2,R_2\to R_2-R_3$

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

$\Rightarrow 1+4\sin 4\theta +{\cos }^2\theta +{\sin }^2\theta =0$
$\Rightarrow \sin 4\theta =-\frac{1}{2}$
$\Rightarrow \sin 4\theta =\sin (\pi +\frac{\pi }{6})or\sin 4\theta =\sin (2\pi -\frac{\pi }{6})$

$\Rightarrow 4\theta =\frac{7\pi }{6}$ or $4\theta =\frac{11\pi }{6}$

$\Rightarrow \theta =\frac{7\pi }{24}\text{or}\frac{11\pi }{24}$