Question

The number of distinct real roots of $\left|\begin{array}{ccc}\sin x& \cos x& \cos x\\ \cos x& \sin x& \cos x\\ \cos x& \cos x& \sin x\end{array}\right|$ in the interval $[\frac{-\pi }{4},\frac{\pi }{4}]$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

Evaluating the determinant

$△=\sin x({\sin }^{2}x-{\cos }^{2}x)-\cos \left(x\right)(\cos x.\sin x-{\cos }^{2}x)+\cos x({\cos }^{2}x-\cos x.\sin x)$

$={\sin }^{3}x-\sin x.{\cos }^{2}x-{\cos }^{2}x\sin x+{\cos }^{3}x+{\cos }^{3}x-{\cos }^{2}x.\sin x$

$={\sin }^{3}x+2{\cos }^{3}x-3{\cos }^{2}x\sin x$

$△=0$ implies

${\sin }^{3}x+2{\cos }^{3}x-3{\cos }^{2}x\sin x=0$

$⟹$$\sin \left(x\right)({\sin }^{2}x-{\cos }^{2}x)+2{\cos }^{2}x(\cos x-\sin x)=0$

$⟹$$-\sin x\left[\right(\cos x+\sin x\left)\right(\cos x-\sin x\left)\right]+2{\cos }^{2}x(\cos x-\sin x)=0$

$⟹$$(\cos x-\sin x)(2{\cos }^{2}x-\sin x(\cos x+\sin x\left)\right)=0$

$⟹$$(\cos x-\sin x)\left(2{\cos }^2\right(x)-{\sin }^{2}x-\sin x\cos x)=0$

$⟹$$(\cos x-\sin x)\left(2{\cos }^2\right(x)+{\cos }^2(x)-{\cos }^2(x)-{\sin }^{2}x-\sin x\cos x)=0$

$⟹$$(\cos x-\sin x)\left(3{\cos }^2\right(x)-\sin x\cos x-1)=0$

Therefore

$\cos x-\sin x=0$ or $\cos x=\sin x$ or $x=\frac{\pi }{4}$ since $x\in [\frac{-\pi }{4},\frac{\pi }{4}]$

And, $3{\cos }^{2}x-\sin x\cos x-1=0$

$⟹$$3{\cos }^{2}x-1=\sin x\cos x$

$⟹$$9{\cos }^{4}x-6{\cos }^{2}x+1={\cos }^{2}x.{\sin }^{2}x$ ... [Squaring both sides]

$⟹$$9{\cos }^{4}x-6{\cos }^{2}x+1=\left({\cos }^{2}x\right)(1-{\cos }^{2}x)$

$⟹$$9{\cos }^{4}x-6{\cos }^{2}x+1=-{\cos }^{4}x+{\cos }^{2}x$

$⟹$$10{\cos }^{4}x-7{\cos }^{2}x+1=0$

Therefore,

${\cos }^2\left(x\right)=\frac{1}{2}$ and ${\cos }^2\left(x\right)=\frac{1}{5}$

$\cos \left(x\right)=\frac{\pm 1}{\sqrt{2}}$ and $\cos \left(x\right)=\left(\frac{\pm 1}{\sqrt{5}}\right)$

$x=\pm \frac{\pi }{4}$ and $x=\pm {63}^0$

Therefore the solutions for the above equation in the interval $x\in [\frac{-\pi }{4},\frac{\pi }{4}]$ are

$-{45}^0,{45}^0$

Or $\pm \frac{\pi }{4}$.

Hence only 2 solutions.