The number of distinct real roots of the equation, \begin{vmatrix} cosx & sinx & sinx \ sinx & cosx & sinx \ sinx & sinx & cosx \end{vmatrix} = 0 in the interval [-π/4, π/4] is/are :

The given determinant is :
\begin{vmatrix} cosx & sinx & sinx \ sinx & cosx & sinx \ sinx & sinx & cosx \end{vmatrix} = 0
Taking cos3x common in the above determinant, we get,
cos3x \begin{vmatrix} 1 & tanx & tanx \ 1 & tanx & tanx \ 1 & tanx & tanx \end{vmatrix} = 0
cos3x[1(1-tan^2x)-tanx(tanx-tan^2x)+tanx(tan^2x-tanx)]=0
cos3x[1-tan^2x+2tan^3x]=0
We notice that if cos3x=0 → x=π/2
This is not satisfied by the given interval
Thus, x=π/2 is not a solution.
Now, 1-tan^2x+2tan^3x=0
It is obvious from the above equation that tanx=1 is a solution.
Thus, x=π/4 is a solution.
We just found out that (tanx-1) is a factor of the polynomial 1-tan^2x+2tan^3x=0
Thus, dividing the polynomial by (tanx-1), we get the quotient as 2tan^2x-tanx-1 and remainder 0
Thus, this can be written as (2tanx+1)(tanx-1)=0
Or, tanx=-1/2 or tanx=1
We already know that tanx=1 is a solution to the above equation.
Thus, tanx=-1/2
Or, x=tan^-1(-1/2)
This also lies in the interval [-π/4, π/4]
Thus there are two distinct real roots to the above equation.