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 interval $[-\frac{\pi }{4},\frac{\pi }{4}],$ is

Answer 1

$\left|\begin{array}{ccc}\cos x& \sin x& \sin x\\ \sin x& \cos x& \sin x\\ \sin x& \sin x& \cos x\end{array}\right|=0$
$C_1\to C_1-C_3,C_2\to C_2-C_3$

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

$\Rightarrow (\cos x-\sin x{)}^2\left|\begin{array}{ccc}1& 0& \sin x\\ 0& 1& \sin x\\ -1& -1& \cos x\end{array}\right|=0$
$R_3\to R_3+R_2$

$\Rightarrow (\cos x-\sin x{)}^2\left|\begin{array}{ccc}1& 0& \sin x\\ 0& 1& \sin x\\ -1& 0& \cos x+\sin x\end{array}\right|=0$

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

Therefore, $2\sin x+\cos x=0$ or $\sin x=\cos x$

For $2\sin x+\cos x=0,\tan x=-\frac{1}{2};$ therefore, one solution in

$x\in [-\frac{\pi }{4},\frac{\pi }{4}]$

For $\sin x=\cos x$, one solution in $x\in [-\frac{\pi }{4},\frac{\pi }{4}]$

Therefore, the total number of solutions is $2.$