Q74 / 160 · WBJEE 2015 · Mathematics

The number of distinct real roots of $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is

  • A
    0
  • B
    2
  • Correct answer
    C
    1
  • D
    >2

Solution

Given, $\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ $\sin x\left(\sin ^{2} x-\cos ^{2} x\right)-\cos x\left(\sin x \cos x-\cos ^{2} x \right ) + \cos x \left(\cos ^{2} x-\sin x \cos x\right)=0$ $\Rightarrow \sin x(\sin x-\cos x)(\sin x+\cos x)$ $-\cos ^{2} x(\sin x-\cos x)-\cos ^{2} x$ $(\sin x-\cos x)=0$ $\Rightarrow(\sin x-\cos x)\left(\sin ^{2} x+\sin x \cos x-2 \cos ^{2} x\right)=0$ $\Rightarrow(\sin x-\cos x)\left(\sin ^{2} x+2 \sin x\right.\left.\cos x-\sin x \cos x-2 \cos ^{2} x\right)=0$ $\Rightarrow(\sin x-\cos x)^{2}(\sin x+2 \cos x)=0$ $\Rightarrow \sin x-\cos x=0$ or $\sin x+2 \cos x=0$ $\Rightarrow \quad \tan x=1$ or $\tan x=-2$ But $\quad x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ $\Rightarrow$ Only one solution i.e. $x=\frac{\pi}{4}$

Hence, the correct answer is option: 'C'.

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