Question

The value of $\theta \in \left[0,\frac{\mathrm{\pi }}{2}\right]$ and satisfying the equation $\left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\sin 4\theta \\ {\cos }^2\theta & 1+{\sin }^2\theta & 4\sin 4\theta \\ {\cos }^2\theta & {\sin }^2\theta & 1+4\sin 4\theta \end{array}\right|$ is

• A. $\frac{11\mathrm{\pi }}{24}$
• B. $\frac{17\mathrm{\pi }}{24}$
• C. $\frac{5\mathrm{\pi }}{24}$
• D. $\frac{\mathrm{\pi }}{24}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Solve the given determinant

Given,

$$\begin{gathered} \left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\sin 4\theta \\ {\cos }^2\theta & 1+{\sin }^2\theta & 4\sin 4\theta \\ {\cos }^2\theta & {\sin }^2\theta & 1+4\sin 4\theta \end{array}\right|=0 \\ {R}_3\to R_3-R_2,R_2\to R_2-R_1 \\ \left|\begin{array}{ccc}1+{\cos }^2\theta & {\sin }^2\theta & 4\sin 4\theta \\ -1& 1& 0\\ 0& -1& 1\end{array}\right|=0 \\ {C}_1\to C_1+C_2 \\ \left|\begin{array}{ccc}1+{\cos }^2\theta +{\sin }^2\theta & {\sin }^2\theta & 4\sin 4\theta \\ -1+1& 1& 0\\ 0-1& -1& 1\end{array}\right|=0 \\ \left|\begin{array}{ccc}2& {\sin }^2\theta & 4\sin 4\theta \\ 0& 1& 0\\ -1& -1& 1\end{array}\right|=0 \end{gathered}$$

Now, if we put $4\sin 2\theta =-2$ then the value of the determinant will be $0$

$$\begin{gathered} \therefore 4\sin 4\theta =-2 \\ 4\sin 4\theta +2=0 \\ \sin 4\theta =-\frac{1}{2}=-\sin 30°=-\sin \frac{\mathrm{\pi }}{6} \\ \therefore \sin 4\theta =\sin \left(-\frac{\mathrm{\pi }}{6}\right)\left[\because -\mathrm{sin\theta }=\sin (-\theta )\right] \\ \therefore 4\theta =n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\left(-\frac{\mathrm{\pi }}{6}\right) \end{gathered}$$

$\therefore$The $\theta$ will be $\frac{7\mathrm{\pi }}{24}$ and $\frac{11\mathrm{\pi }}{24}.$

Hence, the correct answer is option (A).