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The number of distinct real roots of $\begin{vmatrix}\sin x&\cos x&\cos x\\\cos x&\sin x&\cos x\\ \cos x&\cos x&\sin x\end{vmatrix}=0$ in the interval $-\large\frac{\pi}{4}$$\leq x\leq \large\frac{\pi}{4}$ is


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To simplify let $\sin x=\alpha,\cos x=b$ the equation becomes
$\begin{vmatrix}a &b &b\\b&a&b\\b&b&a\end{vmatrix}=0$
Operating $C_2-C_1,C_3-C_2$ we get,
$\begin{vmatrix}a &b-a&o\\b&a-b&b-a\\b&0&a-b\end{vmatrix}=0$
$\Rightarrow a(a-b)^2-(b-a)[b(a-b)-b(b-a)]=0$
$\Rightarrow a(a-b)^2-2b(b-a)(a-b)=0$
$\Rightarrow (a-b)^2(a-2b)=0$
$\Rightarrow (a=b)$ or $a=2b$
$\large\frac{a}{b}$$=1$ , $\large\frac{a}{b}$$=2$
$\tan x=1$ or $\tan x=2$
But we have $-\large\frac{\pi}{4}$$\leq x\leq \large\frac{\pi}{4}$
$\Rightarrow \tan\large\frac{-\pi}{4}$$\leq \tan x\leq \tan\large\frac{\pi}{4}$
$\Rightarrow -1\leq \tan x\leq 1$
$\therefore \tan x=1$
Only one real root is there.
Hence (c) is the right option.
answered Oct 14, 2013 by sreemathi.v

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