The number of distinct real roots of [| pmatrix x & x & x x & x & x x & x & x pmatrix | = …

Question

The number of distinct real roots

of $$\left| {\begin{pmatrix} {\sin x} & {\cos x} & {\cos x}  \
   {\cos x} & {\sin x} & {\cos x}  \
   {\cos x} & {\cos x} & {\sin x}  \ \end{pmatrix} } \right| = 0$$ in the interval $$ - {\pi  \over 4} \le x \le {\pi  \over 4}$$ is :

Accepted Answer

$$\left| {\begin{pmatrix} {\sin x} & {\cos x} & {\cos x}  \
   {\cos x} & {\sin x} & {\cos x}  \
   {\cos x} & {\cos x} & {\sin x}  \

\end{pmatrix} } \right| = 0, - {\pi  \over 4} \le x \le {\pi  \over 4}$$

Apply : $${R_1} \to {R_1} - {R_2}$$ \& $${R_2} \to {R_2} - {R_3}$$

$$ \Rightarrow $$ $$\left| {\begin{pmatrix} {\sin x - \cos x} & {\cos x - \sin x} & 0  \
   0 & {\sin x - \cos x} & {\cos x - \sin x}  \
   {\cos x} & {\cos x} & {\sin x}  \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow $$ $${(\sin x - \cos x)^2}\left| {\begin{pmatrix} 1 & { - 1} & 0  \
   0 & 1 & { - 1}  \
   {\cos x} & {\cos x} & {\sin x}  \

\end{pmatrix} } \right| = 0$$

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

$$\therefore$$ $$x = {\pi  \over 4}$$

Solutions · 1