Determinants — Class 12 Maths Solution

We have,

$\left| {\begin{array}{llllllllllllllllllll}{\sin x}&{\cos x}&{\cos x}\\{\cos x}&{\sin x}&{\cos x}\\{\cos x}&{\cos x}&{\sin x}\end{array}} \right| = 0$

Applying ${C_1} \to {C_1} + {C_2} + {C_3}$,

$\left| {\begin{array}{llllllllllllllllllll}{2\cos x + \sin x}&{\cos x}&{\cos x}\\{2\cos x + \sin x}&{\sin x}&{\cos x}\\{2\cos x + \sin x}&{\cos x}&{\sin x}\end{array}} \right| = 0$

On taking $(2\cos x + \sin x)$ common from ${C_1}$,

we get
$\Rightarrow$ $(2\cos x + \sin x)\left| {\begin{array}{cccccccccccccccccccc}1&{\cos x}&{\cos x}\\1&{\sin x}&{\cos x}\\1&{\cos x}&{\sin x}\end{array}} \right| = 0$

$\Rightarrow$ $(2\cos x + \sin x)\left| {\begin{array}{cccccccccccccccccccc}1&{\cos x}&{\cos x}\\0&{\sin x - \cos x}&0\\0&0&{(\sin x - \cos x)}\end{array}} \right| = 0$

[. and ${R_3} \to {R_3} - {R_1}$]
Expanding along ${C_1}$,

$(2\cos x + \sin x)\left[ {1 \cdot {{(\sin x - \cos x)}^2}} \right] = 0$

$\Rightarrow$ $(2\cos x + \sin x){(\sin x - \cos x)^2} = 0$

Either $2\cos x = - \sin x$

$\Rightarrow$ $\cos x = - \frac{1}{2}\sin x$
$\Rightarrow$ $\tan x = - 2$

…….(i)
But here for $- \frac{\pi }{4} \le x \le \frac{\pi }{4}$,

we get $- 1 \le \tan x \le 1$ so,

no Solution possible and for ${(\sin x - \cos x)^2} = 0,\sin x = \cos x$

$\Rightarrow$ $\tan x = 1 = \tan \frac{\pi }{4}$
$x = \frac{\pi }{4}$

So, only one distinct real root exist.