Question

The values of $\theta$ lying between 0 and $\frac{\pi }{2}$ 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)

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

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

$C_1\to C_1+C_2$

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

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

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

$R_2:R_2\to R_2-R_1$ $R_3\to R_3-R_1$

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

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

Expand

$\Rightarrow 0+\left(1\right)[2-(-1\left)\right(4\sin 4\theta \left)\right]=0$

$\Rightarrow 2+4\sin 4\theta =0$

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

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

$\Rightarrow 4\theta =\pi +\frac{\pi }{6},4\theta =2\pi -\frac{\pi }{6}$ [$\because$ only in $3^{rd}$ and $4^{th}$ quadrants $\sin$ is negative]

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

$\theta =\frac{7\pi }{24},\frac{11\pi }{24}$