Question

In the interval $ \left[- \frac{\pi}{4}, \frac{\pi}{4}\right] $ , the number of real solutions of the equations
$\begin{vmatrix}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{vmatrix}=0$ is

• A. 0
• B. 2
• C. 1
• D. 3

Answer 1

Answer: (C)

Since $\begin{vmatrix}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{vmatrix}=0$
$ \Rightarrow \sin x\left(\sin ^{2} x-\cos ^{2} x\right)-\cos x(\cos x \sin x \left.-\cos ^{2} x\right)+\cos x\left(\cos ^{2} x-\sin x \cos x\right)=0$
$\Rightarrow \sin x\left(\sin ^{2} x-\cos ^{2} x\right)-2 \cos ^{2}x(\sin x-\cos x)=0$
$\Rightarrow (\sin x-\cos x)[\sin x(\sin x+\cos x) \left.-2 \cos ^{2} x\right]=0$
$ \Rightarrow (\sin x-\cos x)[\left(\sin ^{2} x-\cos ^{2} x\right) \left.+\left(\sin x \cos x-\cos ^{2} x\right)\right]=0 $
$\Rightarrow (\sin x-\cos x)^{2}[\sin x+\cos x+\cos x]=0$
$\Rightarrow (\sin x-\cos x)^{2}(\sin x+2 \cos x)=0$
$\Rightarrow $Either $(\sin x-\cos x)^{2}=0$
or $(\sin x+2 \cos x)=0$
$\Rightarrow $ Either $\tan x=1$ or $\tan x=-2$
$\Rightarrow $ Either $x=\frac{\pi}{4}$ or $\tan x=-2$
As $x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], \tan x \in[-1,1]$
Hence, real solution is only $x=\frac{\pi}{4}$