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/are :

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

Answer: (B)

$2$

The given determinant is : $\left|\begin{array}{ccc}\cos x& \sin x& \sin x\\ \sin x& \cos x& \sin x\\ \sin x& \sin x& \cos x\end{array}\right|=0$

Taking ${\cos }^{3}x$ common in the above determinant, we get,

${\cos }^{3}x\left|\begin{array}{ccc}1& tanx& tanx\\ tanx& 1& tanx\\ tanx& tanx& 1\end{array}\right|=0$

${\cos }^{3}x\left[1\right(1-{\tan }^{2}x)-tanx(tanx-ta{n}^{2}x)+tanx(ta{n}^{2}x-tanx\left)\right]=0$

${\cos }^{3}x[1-3ta{n}^{2}x+2ta{n}^{3}x]=0$

We notice that if ${\cos }^{3}x=0\to x=\frac{\pi }{2}$

This is not satisfied by the given interval

Thus, $x=\frac{\pi }{2}$ is not a solution.

Now, $1-3ta{n}^{2}x+2ta{n}^{3}x=0$

It is obvious from the above equation that $tanx=1$ is a solution.

Thus, $x=\frac{\pi }{4}$ is a solution.

We just found out that $(tanx-1)$ is a factor of the polynomial $1-3ta{n}^{2}x+2ta{n}^{3}x=0$

Thus, dividing the polynomial by $(tanx-1)$, we get the quotient as $2tan{x}^2-tanx-1$ and remainder $0$

Thus, this can be written as $(2tanx+1)(tanx-1)=0$

Or, $tanx=\frac{-1}{2}$ or $tanx=1$

We already know that $tanx=1$ is a solution to the above equation.

Thus, $tanx=\frac{-1}{2}$

Or, $x=ta{n}^{-1}\left(\frac{-1}{2}\right)$

This also lies in the interval $[\frac{-\pi }{4},\frac{\pi }{4}]$
Thus there are two distinct real roots to the above equation.