"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

Question

"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"

Accepted Answer

"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

$1(1 + {\cos ^2}x + 4\sin 2x) - {\sin ^2}x( - 1) = 0$

$\Rightarrow 2 + 4\sin 2x = 0$

$\Rightarrow \sin 2x = {{ - 1} \over 2}$

$\Rightarrow 2x = n\pi + {( - 1)^n}\left( {{{ - \pi } \over 6}} \right),n \in Z$

$\therefore$ $2x = {{7\pi } \over 6},{{11\pi } \over 6}$

$\Rightarrow x = {{7\pi } \over {12}},{{11\pi } \over 2}$"

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

Solution

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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

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