Question

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

• A. 0
• B. 2
• C. 1
• D. 3

Answer 1

Answer: (C)

Since $\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|=0$
$\Rightarrow \sin x\left({\sin }^{2}x-{\cos }^{2}x\right)-\cos x(\cos x\sin x-{\cos }^{2}x)+\cos x\left({\cos }^{2}x-\sin x\cos x\right)=0$
$\Rightarrow \sin x\left({\sin }^{2}x-{\cos }^{2}x\right)-2{\cos }^{2}x(\sin x-\cos x)=0$
$\Rightarrow (\sin x-\cos x)[\sin x(\sin x+\cos x)-2{\cos }^{2}x]=0$
$\Rightarrow (\sin x-\cos x)[\left({\sin }^{2}x-{\cos }^{2}x\right)+\left(\sin x\cos x-{\cos }^{2}x\right)]=0$
$\Rightarrow (\sin x-\cos x{)}^2[\sin x+\cos x+\cos x]=0$
$\Rightarrow (\sin x-\cos x{)}^2(\sin x+2\cos x)=0$
$\Rightarrow$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 $x=\frac{\pi }{4}$