The value of θ lying between −π4 and π2 and 0≤A≤π2 and satisfying the equation 1+sin2⁡Acos2⁡A2sin⁡4θsin2⁡A1+cos2⁡A2sin⁡4θsin2⁡Acos2⁡A1+2sin⁡4θ=0 ,are

∵ 1+sin2⁡Acos2⁡A2sin⁡4θsin2⁡A1+cos2⁡A2sin⁡4θsin2⁡Acos2⁡A1+2sin⁡4θ=0Applying R2→R2−R1 and R3→R3−R1 then1+sin2⁡Acos2⁡A2sin⁡4θ−110−101=0Applying C1→C1+C2, then2 cos2⁡A2sin⁡4θ ⋮ 0…1…0 ⋮ −1 0 1=0⇒ 1(2+2sin⁡4θ)=0∴ sin⁡4θ=−1⇒ 4θ=(2n−1)π2⇒θ=(2n−1)π8For n = 0, 2, then θ =−π8,3π8 and A∈R.

The value of θ lying between −π4 and π2 and 0≤A≤π2 and satisfying the equation 1+sin2⁡Acos2⁡A2sin⁡4θsin2⁡A1+cos2⁡A2sin⁡4θsin2⁡Acos2⁡A1+2sin⁡4θ=0 ,are

∵ 1+sin2⁡Acos2⁡A2sin⁡4θsin2⁡A1+cos2⁡A2sin⁡4θsin2⁡Acos2⁡A1+2sin⁡4θ=0Applying R2→R2−R1 and R3→R3−R1 then1+sin2⁡Acos2⁡A2sin⁡4θ−110−101=0Applying C1→C1+C2, then2 cos2⁡A2sin⁡4θ ⋮ 0…1…0 ⋮ −1 0 1=0⇒ 1(2+2sin⁡4θ)=0∴ sin⁡4θ=−1⇒ 4θ=(2n−1)π2⇒θ=(2n−1)π8For n = 0, 2, then θ =−π8,3π8 and A∈R.

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The value of $θ$ lying between $- \frac{π}{4} and \frac{π}{2}$ and $0 \leq A \leq \frac{π}{2}$ and satisfying the equation
$|\begin{array}{ccc} 1 + \sin^{2} A & \cos^{2} A & 2 \sin 4 θ \\ \sin^{2} A & 1 + \cos^{2} A & 2 \sin 4 θ \\ \sin^{2} A & \cos^{2} A & 1 + 2 \sin 4 θ \end{array}| = 0$ ,are

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$A = \frac{π}{4}, θ = - \frac{π}{8}$

$A = \frac{3 π}{8} = θ$

$A = \frac{π}{5}, θ = - \frac{π}{8}$

$A = \frac{π}{6}, θ = \frac{3 π}{8}$

answer is A, B, C, D.

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Detailed Solution

$∵ |\begin{array}{ccc} 1 + \sin^{2} A & \cos^{2} A & 2 \sin 4 θ \\ \sin^{2} A & 1 + \cos^{2} A & 2 \sin 4 θ \\ \sin^{2} A & \cos^{2} A & 1 + 2 \sin 4 θ \end{array}| = 0$

Applying $R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}$ then

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

Applying $C_{1} \to C_{1} + C_{2}$ , then

$\begin{array}{l} |\begin{array}{cccc} 2 & \cos^{2} A & 2 \sin 4 θ \\ ⋮ \\ 0 & \dots & 1 & \dots & 0 \\ ⋮ \\ - 1 & 0 & 1 \end{array}| = 0 \\ \Rightarrow 1 (2 + 2 \sin 4 θ) = 0 \\ ∴ \sin 4 θ = - 1 \end{array}$