JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 2)
The solutions of the equation $$\left| {\matrix{
{1 + {{\sin }^2}x} & {{{\sin }^2}x} & {{{\sin }^2}x} \cr
{{{\cos }^2}x} & {1 + {{\cos }^2}x} & {{{\cos }^2}x} \cr
{4\sin 2x} & {4\sin 2x} & {1 + 4\sin 2x} \cr
} } \right| = 0,(0 < x < \pi )$$, are
$${\pi \over {12}},{\pi \over 6}$$
$${\pi \over 6},{{5\pi } \over 6}$$
$${{5\pi } \over {12}},{{7\pi } \over {12}}$$
$${{7\pi } \over {12}},{{11\pi } \over {12}}$$
Explanation
By using C1 $$ \to $$ C1 $$-$$ C2 and C3 $$ \to $$ C3 $$-$$ C2 we get
$$\left| {\matrix{ 1 & {{{\sin }^2}x} & 0 \cr { - 1} & {1 + {{\cos }^2}x} & { - 1} \cr 0 & {4\sin 2x} & 1 \cr } } \right| = 0$$
Expanding by R1 we get
$$1(1 + {\cos ^2}x + 4\sin 2x) - {\sin ^2}x( - 1) = 0$$
$$ \Rightarrow 2 + 4\sin 2x = 0$$
$$ \Rightarrow \sin 2x = {{ - 1} \over 2}$$
$$ \Rightarrow 2x = n\pi + {( - 1)^n}\left( {{{ - \pi } \over 6}} \right),n \in Z$$
$$ \therefore $$ $$2x = {{7\pi } \over 6},{{11\pi } \over 6} $$
$$\Rightarrow x = {{7\pi } \over {12}},{{11\pi } \over 2}$$
$$\left| {\matrix{ 1 & {{{\sin }^2}x} & 0 \cr { - 1} & {1 + {{\cos }^2}x} & { - 1} \cr 0 & {4\sin 2x} & 1 \cr } } \right| = 0$$
Expanding by R1 we get
$$1(1 + {\cos ^2}x + 4\sin 2x) - {\sin ^2}x( - 1) = 0$$
$$ \Rightarrow 2 + 4\sin 2x = 0$$
$$ \Rightarrow \sin 2x = {{ - 1} \over 2}$$
$$ \Rightarrow 2x = n\pi + {( - 1)^n}\left( {{{ - \pi } \over 6}} \right),n \in Z$$
$$ \therefore $$ $$2x = {{7\pi } \over 6},{{11\pi } \over 6} $$
$$\Rightarrow x = {{7\pi } \over {12}},{{11\pi } \over 2}$$
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