Question

The number of distinct real roots of the equation, $\left| {\begin{array}{*{20}{c}}
  {\cos x}&{\sin x}&{\sin x} \\
  {\sin x}&{\cos x}&{\sin x} \\
  {\sin x}&{\sin x}&{\cos x}
\end{array}} \right| = 0$ in the interval $\left[ { - \dfrac{\pi }{4},\dfrac{\pi }{4}} \right]$ is/are:
A. $3$
B. $2$
C. $1$
D. $4$

Answer 1

Hint:At first, use the row or column operations to simplify the determinant as much as you can. Once it is simplified, open the determinant and equate it to $0$. Find the value of $x$ such that the equation satisfies. Count the number of values.

Complete step-by-step answer:
$\left| {\begin{array}{*{20}{c}}
  {\cos x}&{\sin x}&{\sin x} \\
  {\sin x}&{\cos x}&{\sin x} \\
  {\sin x}&{\sin x}&{\cos x}
\end{array}} \right| = 0$

Using column operator, ${C_1} \to {C_1} - {C_2}$
$\left| {\begin{array}{*{20}{c}}
  {\cos x - \sin x}&{\sin x}&{\sin x} \\
  {\sin x - \cos x}&{\cos x}&{\sin x} \\
  0&{\sin x}&{\cos x}
\end{array}} \right| = 0$

Using column operator, ${C_2} \to {C_2} - {C_3}$
$\left| {\begin{array}{*{20}{c}}
  {\cos x - \sin x}&0&{\sin x} \\
  {\sin x - \cos x}&{\cos x - \sin x}&{\sin x} \\
  0&{\sin x - \cos x}&{\cos x}
\end{array}} \right| = 0$

Taking $\sin x - \cos x$ common from ${C_1}$
$(\sin x - \cos x)$$\left| {\begin{array}{*{20}{c}}
  { - 1}&0&{\sin x} \\
  1&{\cos x - \sin x}&{\sin x} \\
  0&{\sin x - \cos x}&{\cos x}
\end{array}} \right| = 0$

Taking $(\sin x - \cos x)$ from ${C_2}$
${(\sin x - \cos x)^2}$$\left| {\begin{array}{*{20}{c}}
  { - 1}&0&{\sin x} \\
  1&{ - 1}&{\sin x} \\
  0&1&{\cos x}
\end{array}} \right| = 0$

Now the determinant is simplified.
So, we will open it.
${(\sin x - \cos x)^2}$$\left[ { - 1( - \cos x - \sin x) + 0 + \sin x} \right] = 0$
${(\sin x - \cos x)^2}$$\left[ {\cos x + 2\sin x} \right] = 0$

At least one of the terms must be $0$ in order to satisfy the equation.
Hence, ${(\sin x - \cos x)^2}$$ = 0$ or $\left[ {\cos x + 2\sin x} \right] = 0$
$\sin x = \cos x$ or $2\sin x = -\cos x$
We know $\dfrac{\sin x}{\cos } = \tan x$
Simplifying we get $\tan x= 1$ or $\tan x = - \dfrac{1}{2}$
$x = \dfrac{\pi }{4}$ or $x = {\tan ^{ - 1}}( - \dfrac{1}{2})$

Now range of $x$ is $\left[ { - \dfrac{\pi }{4},\dfrac{\pi }{4}} \right]$
Hence possible values of $x$ are $\dfrac{\pi }{4}$and ${\tan ^{ - 1}}( - \dfrac{1}{2})$
Hence there are two distinct real roots.

So, the correct answer is “Option B”.

Note:Students should be very careful about the signs while performing row or column operation.We have to carefully analyse which operation should be performed .Students should remember trigonometric ratios and method of calculating the determinant value for solving these types of problems.