JEE MAIN - Mathematics (2019 - 12th April Evening Slot - No. 11)
A value of $$\theta \in \left( {0,{\pi \over 3}} \right)$$, for which
$$\left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$$, is :
$$\left| {\matrix{ {1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$$, is :
$${\pi \over {18}}$$
$${\pi \over {9}}$$
$${{7\pi } \over {24}}$$
$${{7\pi } \over {36}}$$
Explanation
$$\left| {\matrix{
{1 + {{\cos }^2}\theta } & {{{\sin }^2}\theta } & {4\cos 6\theta } \cr
{{{\cos }^2}\theta } & {1 + {{\sin }^2}\theta } & {4\cos 6\theta } \cr
{{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr
} } \right| = 0$$
R1 $$ \to $$ R1 - R2, R2 $$ \to $$ R2 - R3
$$ \Rightarrow \left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$$
C2 $$ \to $$ C2 + C1
$$ \Rightarrow \left| {\matrix{ 1 & 0 & 0 \cr 0 & 1 & { - 1} \cr {{{\cos }^2}\theta } & 1 & {1 + 4\cos 6\theta } \cr } } \right| = 0$$
$$ \Rightarrow 1 + 4\cos 6\theta + 1 = 0$$
$$ \Rightarrow 2\cos 6\theta = - 1 \Rightarrow \cos 6\theta = - {1 \over 2}$$ = $$\cos {{2\pi } \over 3}$$
$$ \Rightarrow 6\theta = 2n\pi \pm {{2\pi } \over 3}$$
$$ \Rightarrow \theta = {{n\pi } \over 3} \pm {\pi \over 9}\,\,\,n \in 1$$
$$ \Rightarrow \theta = {\pi \over 9},{{2\pi } \over 9},{{4\pi } \over 9}$$
R1 $$ \to $$ R1 - R2, R2 $$ \to $$ R2 - R3
$$ \Rightarrow \left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr {{{\cos }^2}\theta } & {{{\sin }^2}\theta } & {1 + 4\cos 6\theta } \cr } } \right| = 0$$
C2 $$ \to $$ C2 + C1
$$ \Rightarrow \left| {\matrix{ 1 & 0 & 0 \cr 0 & 1 & { - 1} \cr {{{\cos }^2}\theta } & 1 & {1 + 4\cos 6\theta } \cr } } \right| = 0$$
$$ \Rightarrow 1 + 4\cos 6\theta + 1 = 0$$
$$ \Rightarrow 2\cos 6\theta = - 1 \Rightarrow \cos 6\theta = - {1 \over 2}$$ = $$\cos {{2\pi } \over 3}$$
$$ \Rightarrow 6\theta = 2n\pi \pm {{2\pi } \over 3}$$
$$ \Rightarrow \theta = {{n\pi } \over 3} \pm {\pi \over 9}\,\,\,n \in 1$$
$$ \Rightarrow \theta = {\pi \over 9},{{2\pi } \over 9},{{4\pi } \over 9}$$
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