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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 $\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]$ is
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$\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|$ in the interval $\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]$ is
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The correct answer is:
1
$\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$
Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ $\left|\begin{array}{lll}2 \cos x+\sin x & \cos x & \cos x \\ 2 \cos x+\sin x & \sin x & \cos x \\ 2 \cos x+\sin x & \cos x & \sin x\end{array}\right|=0$
$\Rightarrow(2 \cos x+\sin x)\left|\begin{array}{lll}1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x\end{array}\right|=0$
[Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we get $(2 \cos x+\sin x)\left|\begin{array}{ccc}1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 0 & 0 & \sin x-\cos x\end{array}\right|=0$
$\Rightarrow(2 \cos x+\sin x)\left[1 \cdot(\sin -\cos x)^{2}\right]=0$
$\Rightarrow(2 \cos x+\sin x)(\sin x-\cos x)^{2}=0$
$\Rightarrow 2 \cos x=-\sin x$ or $\sin x=\cos x$
$\Rightarrow \tan x=-2$ which is not possible as for
$-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$, we get $-1 \leq \tan x \leq 1$
or $\quad \tan x=1$
So, $x=\frac{\pi}{4}$. Hence, only one real root exist.
Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ $\left|\begin{array}{lll}2 \cos x+\sin x & \cos x & \cos x \\ 2 \cos x+\sin x & \sin x & \cos x \\ 2 \cos x+\sin x & \cos x & \sin x\end{array}\right|=0$
$\Rightarrow(2 \cos x+\sin x)\left|\begin{array}{lll}1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x\end{array}\right|=0$
[Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we get $(2 \cos x+\sin x)\left|\begin{array}{ccc}1 & \cos x & \cos x \\ 0 & \sin x-\cos x & 0 \\ 0 & 0 & \sin x-\cos x\end{array}\right|=0$
$\Rightarrow(2 \cos x+\sin x)\left[1 \cdot(\sin -\cos x)^{2}\right]=0$
$\Rightarrow(2 \cos x+\sin x)(\sin x-\cos x)^{2}=0$
$\Rightarrow 2 \cos x=-\sin x$ or $\sin x=\cos x$
$\Rightarrow \tan x=-2$ which is not possible as for
$-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$, we get $-1 \leq \tan x \leq 1$
or $\quad \tan x=1$
So, $x=\frac{\pi}{4}$. Hence, only one real root exist.
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