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Mathematics Question on Determinants

A value of θ  ∈ (0,π/3)\theta \ \ \in \ (0, \pi /3) for which ∣1+cos2θsin2θ4cos6θ cos2θ1+sin2θ4cos6θ cos2θsin2θ1+4cos6θ∣\begin {vmatrix} 1 + cos^2 \theta & sin^2\theta & 4 cos 6\theta \\\ cos^2 \theta & 1+sin^2 \theta & 4 cos6 \theta \\\ cos^2 \theta & sin^2 \theta & 1+4 cos6 \theta \end {vmatrix} =0 , is :

A

7π24\frac{7\pi}{24}

B

π18\frac{\pi}{18}

C

π9\frac{\pi}{9}

D

7π36\frac{7\pi}{36}

Answer

π9\frac{\pi}{9}

Explanation

Solution

R1⇒R1−R2R_1 \Rightarrow R_1 - R_2
∣1−10 cos2θ1+sin2θ4cos6θ cos2θsin2θ1+4cos6θ∣\begin {vmatrix} 1 & -1 & 0 \\\ cos^2 \theta & 1+sin^2 \theta & 4 cos6 \theta \\\ cos^2 \theta & sin^2 \theta & 1+4 cos6 \theta \end {vmatrix} = 0
R2 → R2−R3R_2 \ \rightarrow \ R_2 - R_3
∣1−10 01−1 cos2θsin2θ1+4cos6θ∣\begin {vmatrix} 1 & -1 & 0 \\\ 0 & 1 & -1 \\\ cos^2 \theta & sin^2\theta & 1+4 cos 6 \theta \end {vmatrix} = 0
⇒  (1+4cos6θ)+sin2θ+1(cos2θ)=0\Rightarrow \ \ (1 + 4 cos6 \theta) + sin^2 \theta + 1 (cos^2 \theta) = 0
   1+2cos6θ=0⇒  cos6θ =−1/2\ \ \ 1 + 2 cos6\theta = 0 \Rightarrow \ \ cos 6 \theta \ = -1/2
6θ =2π3 ⇒  θ=π96\theta \ = \frac{2\pi}{3} \ \Rightarrow \ \ \theta = \frac{\pi}{9}