If A=1+cos2⁡θsin2⁡θ4cos⁡6θcos2⁡θ1+sin2⁡θ4cos⁡6θcos2⁡θsin2⁡θ1+4cos⁡6θ is a singular matrix and 3θ∈(0,π) then θ=

|A|=0⇒2+4cos⁡6θ=0⇒cos⁡6θ=−12 ⇒6θ=2π3⇒θ=π9

If A=1+cos2⁡θsin2⁡θ4cos⁡6θcos2⁡θ1+sin2⁡θ4cos⁡6θcos2⁡θsin2⁡θ1+4cos⁡6θ is a singular matrix and 3θ∈(0,π) then θ=

|A|=0⇒2+4cos⁡6θ=0⇒cos⁡6θ=−12 ⇒6θ=2π3⇒θ=π9

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$If A = [\begin{array}{ccc} 1 + \cos^{2} θ & \sin^{2} θ & 4 \cos 6 θ \\ \cos^{2} θ & 1 + \sin^{2} θ & 4 \cos 6 θ \\ \cos^{2} θ & \sin^{2} θ & 1 + 4 \cos 6 θ \end{array}] is a singular matrix and 3 θ \in (0, π) then θ =$

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$\frac{π}{9}$

$\frac{π}{18}$

$\frac{π}{6}$

$\frac{π}{12}$

answer is A.

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$| A | = 0 \Rightarrow 2 + 4 \cos 6 θ = 0 \Rightarrow \cos 6 θ = - \frac{1}{2} \Rightarrow 6 θ = \frac{2 π}{3} \Rightarrow θ = \frac{π}{9}$

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If A=1+cos2⁡θsin2⁡θ4cos⁡6θcos2⁡θ1+sin2⁡θ4cos⁡6θcos2⁡θsin2⁡θ1+4cos⁡6θ is a singular matrix and 3θ∈(0,π) then θ=

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$If A = [\begin{array}{ccc} 1 + \cos^{2} θ & \sin^{2} θ & 4 \cos 6 θ \\ \cos^{2} θ & 1 + \sin^{2} θ & 4 \cos 6 θ \\ \cos^{2} θ & \sin^{2} θ & 1 + 4 \cos 6 θ \end{array}] is a singular matrix and 3 θ \in (0, π) then θ =$