Question

The number of distinct real roots of the equation, $\left|\begin{array}{ccc}\cos x& \sin x& \sin x\\ \sin x& \cos x& \sin x\\ \sin x& \sin x& \cos x\end{array}\right|=0$ in the intervals $[-\frac{\pi }{4},\frac{\pi }{4}]$ is:

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

Let $\Delta =\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|$

Apply $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$

$\Delta =\left|\begin{array}{ccc}\sin x-\cos x& \cos x-\sin x& 0\\ 0& \sin x-\cos x& \cos x-\sin x\\ \cos x& \cos x& \sin x\end{array}\right|$

Let us take $(\sin x-\cos x)$ as a common factor from $R_1$ and $R_2$

$=(\sin x-\cos x)\left|\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ \cos x& \cos x& \sin x\end{array}\right|$

Now expanding along $R_1$ we get,

$\Delta ={(\sin x-\cos x)}^2[1(\sin x+\cos x)-1(0+\cos x)]$

$={(\sin x-\cos x)}^2[\sin x+\cos x-\cos x]$

Given that $\Delta =0$

$\Rightarrow {(\sin x-\cos x)}^2\sin x=0$

$\Rightarrow {(\sin x-\cos x)}^2=0$

$\Rightarrow \sin x-\cos x=0$

$\Rightarrow \sin x=\cos x=\frac{1}{\sqrt{2}}$

This is possible if $x=\frac{\pi }{4}$ and $\sin x=0$ if $x=0$

We know that $\frac{1}{\sqrt{2}}$ is irrational.

Hence $\sin x=0$ is the only distinct real root,between the interval $\frac{-\pi }{4}\le x\le \frac{\pi }{4}$

Hence number of distinct real roots$=1$