Question

The value/s 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

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

Apply $C_1\to C_1+C_2$

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

Now applying $R_2\to R_2-R_{1}and{R}_3\to R_3-R_1$

$△=\left|\begin{array}{ccc}2& co{s}^2\theta & 4sin4\theta \\ 0& 1& 0\\ -1& 0& 1\end{array}\right|=0$

$\Rightarrow 2+4sin4\theta =0$ by expansion with $R_2$

$\Rightarrow sin4\theta =-\frac{1}{2}=-sin\left(\frac{\pi }{6}\right)=sin(-\frac{\pi }{6})$

$\therefore 4\theta =n\pi +(-1{)}^n(-\frac{\pi }{6})\therefore \theta =\frac{6n-(-1{)}^n}{24}\pi$

The value of $\theta$ lying between 0 and $\frac{\pi }{2}are\frac{7\pi }{24}and\frac{11\pi }{24}forn=1and2$