Solve : ; where, .

Engineering

Mathematics

Question

Solve : $|\begin{array}{ccc} 1 + \sin^{2} θ & \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & 1 + \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & \cos^{2} θ & 1 + 4 \sin 4 θ \end{array}| = 0$ ; where, $0 < θ < \frac{π}{2}$ .

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Given $|\begin{array}{ccc} 1 + \sin^{2} θ & \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & 1 + \cos^{2} θ & 4 \sin 4 θ \\ \sin^{2} θ & \cos^{2} θ & 1 + 4 \sin 4 θ \end{array}| = 0$ ; where, $0 < θ < \frac{π}{2} .$
(C₁ → C₁ + C₂)
$\Rightarrow |\begin{array}{ccc} 2 & \cos^{2} θ & 4 \sin 4 θ \\ 2 & 1 + \cos^{2} θ & 4 \sin 4 θ \\ 1 & \cos^{2} θ & 1 + 4 \sin 4 θ \end{array}| = 0$

(R₂ → R₂ – R₁ & R₃ → R₃ – R₁)

\Rightarrow |\begin{array}{ccc} 2 & \cos^{2} θ & 4 \sin 4 θ \\ 0 & 1 & 0 \\ - 1 & 0 & 1 \end{array}| = 0

⇒ 2 + 4sin4θ = 0

\Rightarrow \sin 4 θ = - \frac{1}{2}

\Rightarrow 4 θ = \frac{7 π}{6} \dots [Since o < θ < \frac{π}{2}]

\Rightarrow θ = \frac{7 π}{24}