Question

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

(a) 0
(b) 2
(c) 1
(d) 3

Answer 1

(c) We have,
$\left|\begin{array}{ccc}sinx& cosx& cosx\\ cosx& sinx& cosx\\ cosx& cosx& sinx\end{array}\right|=0$
Applying $C_1\to C_1+C_2+C_3$
$\left|\begin{array}{ccc}2cosx+sinx& cosx& cosx\\ 2cosx+sinx& sinx& cosx\\ 2cosx+sinx& cosx& sinx\end{array}\right|=0$
On taking $(2cosx+sinx)$ common from $C_1$, we get
$\Rightarrow (2cosx+sinx)\left|\begin{array}{ccc}1& cosx& cosx\\ 1& sinx& cosx\\ 1& cosx& sinx\end{array}\right|=0\Rightarrow (2cosx+sinx)\left|\begin{array}{ccc}1& cosx& cosx\\ 0& sinx-cosx& 0\\ 0& 0& (sinx-cosx)\end{array}\right|=0$
$[\because R_2\to R_2-R_{1}and{R}_3\to R_3-R_1]$
Expanding along $C_1$,
$(2cosx+sinx)\left[1.\right(sinx-cosx{)}^2]=0\Rightarrow (2cosx+sinx\left)\right(sinx-cosx{)}^2=0$
Either $2cosx=-sinx$
$\Rightarrow cosx=-\frac{1}{2}sinx$
$\Rightarrow tanx=-2$ ..... (ii)
But here for $-\frac{\pi }{4}\le x\le \frac{\pi }{4},weget-1\le tanx\le 1$ so, no solution possible
and for $(sinx-cosx{)}^2=0,sinx=cosx$
$\Rightarrow tanx=1=tan\frac{\pi }{4}\therefore x=\frac{\pi }{4}$
So, only one distinct real root exist.