Question

The value of $\theta$ laying between $\theta =0$ and $\theta =\pi /2$ and satisfying the equation $\left|\begin{array}{ccc}1+si{n}^2\theta & co{s}^2\theta & 4sin4\theta \\ si{n}^2\theta & 1+co{s}^2\theta & 4sin4\theta \\ si{n}^2\theta & co{s}^2\theta & 1+4sin4\theta \end{array}\right|=0$ 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)

$\frac{7\pi }{24}$
$\frac{11\pi }{24}$

given: $\left|\begin{array}{ccc}1+si{n}^2\theta & co{s}^2\theta & 4sin4\theta \\ si{n}^2\theta & 1+co{s}^2\theta & 4sin4\theta \\ si{n}^2\theta & co{s}^2\theta & 1+4sin4\theta \end{array}\right|=0$

Applying $C_1+C_2$,
$=\left[\begin{array}{ccc}2& co{s}^2\theta & 4sin4\theta \\ 2& 1+co{s}^2\theta & 4sin4\theta \\ 1& co{s}^2\theta & 1+4sin4\theta \end{array}\right]$
Applying $R_1-R_2$ ,
$=\left[\begin{array}{ccc}0& -1& 0\\ 2& 1+co{s}^2\theta & 4sin4\theta \\ 1& co{s}^2\theta & 1+4sin4\theta \end{array}\right]$
Expanding,$\therefore 2+4sin4\theta =0$
Hence, $sin4\theta =\frac{-1}{2}$
Hence, options A and C are correct.