Question

The value of $x$ satisfying the equation $\left|\begin{array}{ccc}\cos 2x& \sin 2x& \sin 2x\\ \sin 2x& \cos 2x& \sin 2x\\ \sin 2x& \sin 2x& \cos 2x\end{array}\right|=0$ and $xϵ[0,\frac{\pi }{4}]$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{16}$
• D. $\frac{\pi }{3}$
• E. $\frac{\pi }{8}$

Answer 1

The correct option is E $\frac{\pi }{8}$

Given, $\left|\begin{array}{ccc}\cos 2x& \sin 2x& \sin 2x\\ \sin 2x& \cos 2x& \sin 2x\\ \sin 2x& \sin 2x& \cos 2x\end{array}\right|=0;xϵ[0,\frac{\pi }{4}]$
Applying operation: $R_1\to R_1+R_2+R_3$, we get
$\left|\begin{array}{ccc}\cos 2x+2\sin 2x& 2\sin 2x+\cos 2x& 2\sin 2x+\cos 2x\\ \sin 2x& \cos 2x& \sin 2x\\ \sin 2x& \sin 2x& \cos 2x\end{array}\right|=0$
$\Rightarrow (2\sin 2x+\cos 2x)\left|\begin{array}{ccc}1& 1& 1\\ \sin 2x& \cos 2x& \sin 2x\\ \sin 2x& \sin 2x& \cos 2x\end{array}\right|=0$
Now, apply operation
$C_2\to C_2-C_1$ and $C_3\to C_3-C_1$, we get
$(2\sin 2x+\cos 2x)$
$\left|\begin{array}{ccc}1& 0& 0\\ \sin 2x& \cos 2x-\sin 2x& 0\\ \sin 2x& 0& \cos 2x-\sin 2x\end{array}\right|=0$
Expanding along $R_1$, we get
$(2\sin 2x+\cos 2x)(\cos 2x-\sin 2x{)}^2=0$
$\Rightarrow 2\sin 2x+\cos 2x=0$
or $\cos 2x-\sin 2x=0$
$\Rightarrow \tan 2x=-\frac{1}{2}$
or $\tan 2x=1=\tan \frac{\pi }{4}$
$\because x\ne \frac{1}{2}{\tan }^{-1}\left(\frac{-1}{2}\right)$
$\therefore x=\frac{\pi }{8}ϵ[0,\frac{\pi }{4}]$.