Question
The number of distinct real roots of \[\left| {\begin{array}{*{20}{c}}
{\sin x}&{\cos x}&{\cos x}\\
{\cos x}&{\sin x}&{\cos x}\\
{\cos x}&{\cos x}&{\sin x}
\end{array}} \right| = 0\] in the interval $$ - \frac{\pi }{4} \leqslant x \leqslant \frac{\pi }{4}$$ is
A.
0
B.
2
C.
1
D.
3
Answer :
1
Solution :
To simplify the det. Let $$\sin x = a;\cos x = b$$ the equation becomes
\[\begin{array}{l}
\left| {\begin{array}{*{20}{c}}
a&b&b\\
b&a&b\\
b&b&a
\end{array}} \right| = 0\,\,{\rm{Operating }}\,\,{C_2} - {C_1};{C_3} - {C_2}\,{\rm{we\, get}}\\
\left| {\begin{array}{*{20}{c}}
a&{b - a}&0\\
b&{a - b}&{b - a}\\
b&0&{a - b}
\end{array}} \right| = 0
\end{array}\]
$$\eqalign{
& \Rightarrow \,\,a{\left( {a - b} \right)^2} - \left( {b - a} \right)\left[ {b\left( {a - b} \right) - b\left( {b - a} \right)} \right] = 0 \cr
& \Rightarrow \,\,a{\left( {a - b} \right)^2} - 2b\left( {b - a} \right)\left( {a - b} \right) = 0 \cr
& \Rightarrow \,\,{\left( {a - b} \right)^2}\left( {a - 2b} \right) = 0 \cr
& \Rightarrow \,\,\left( {a = b} \right){\text{or }}a = 2b \cr
& \Rightarrow \,\,\frac{a}{b} = 1\,{\text{or }}\frac{a}{b} = 2 \cr
& \Rightarrow \,\,\tan x = 1\,\,{\text{or }}\tan x = 2.\,{\text{But we have }} - \frac{\pi }{4} \leqslant x \leqslant \frac{\pi }{4} \cr
& \Rightarrow \,\,\tan \left( { - \frac{\pi }{4}} \right) \leqslant \tan x \leqslant \tan \left( {\frac{\pi }{4}} \right) \cr
& \Rightarrow \,\, - 1 \leqslant \tan x \leqslant 1 \cr
& \therefore \,\,\tan x = 1 \cr
& \Rightarrow \,\,x = \frac{\pi }{4} \cr} $$
∴ Only one real root is there.