Question

The number of distinct real roots of $\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0$ in the interval $-\frac{\pi }{4}\le x\le \frac{\pi }{4}$ is

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

The correct option is A $2$

Given: $\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0$
$R_1\to R_1+R_2+R_3$
$\Rightarrow \left|\begin{array}{ccc}\sin x+2\cos x& \sin x+2\cos x& \sin x+2\cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0$
$\Rightarrow (\sin x+2\cos x)\left|\begin{array}{ccc}1& 1& 1\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0$
$C_1\to C_1-C_3,C_2\to C_2-C_3$
$\Rightarrow (\sin x+2\cos x)\left|\begin{array}{ccc}0& 0& 1\\ 0& \sin x-\cos x& \cos x\\ \cos x-\sin x& \cos x-\sin x& \sin x\end{array}\right|=0$
$\Rightarrow (\sin x+2\cos x)(\sin x-\cos x{)}^2=0$
$\Rightarrow \sin x=\cos x$ or $\sin x=-2\cos x$
$\Rightarrow \tan x=1$ or $\tan x=-2$
In the interval of $[-\frac{\pi }{4},\frac{\pi }{4}],$ we have
$\therefore x=\frac{\pi }{4}$