Question

If $\left|\begin{array}{ccc}1+si{n}^2\theta & co{s}^2\theta & 4sin4\theta \\ si{n}^2\theta & 1+co{s}^2\theta & 4sin4\theta \\ si{n}^2\theta & co{s}^2\theta & 1+4sin4\theta \end{array}\right|=0$ such that $0\le \theta \le \frac{\pi }{2}$ then $\theta$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Operating $R_1-R_3,R_2-R_3$ and then $C_3\to C_3+C_1$ we get

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

On Expanding we get

$1+si{n}^2\theta +4sin4\theta +co{s}^2\theta =1$

$\Rightarrow$ 2 + $4sin4\theta =0\Rightarrow sin4\theta =\frac{-1}{2}=sin\left(\frac{-\pi }{6}\right)$

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

For $0\le \theta \le \frac{\pi }{2}$ we have $\theta =\frac{7\pi }{24},\frac{11\pi }{24}$