A value of theta in (0, pi /3),for which left | vedclass

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(B) Given $	heta \in (0, \frac{\pi}{3})$.Let the determinant be $\Delta = \left| \begin{array}{ccc} 1 + \cos^2 	heta & \sin^2 	heta & 4 \cos 6	het

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$\frac{\pi }{9}$

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(B) Given $	heta \in (0, \frac{\pi}{3})$.Let the determinant be $\Delta = \left| \begin{array}{ccc} 1 + \cos^2 	heta & \sin^2 	heta & 4 \cos 6	heta \ \cos^2 	heta & 1 + \sin^2 	heta & 4 \cos 6	heta \ \cos^2 	heta & \sin^2 	heta & 1 + 4 \cos 6	heta \end{array} ight| = 0$.Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta = \left| \begin{array}{ccc} 1 + \cos^2 	heta & \sin^2 	heta & 4 \cos 6	heta \ -1 & 1 & 0 \ -1 & 0 & 1 \end{array} ight| = 0$.Applying column operation $C_1 	o C_1 + C_2$:$\Delta = \left| \begin{array}{ccc} 1 + \cos^2 	heta + \sin^2 	heta & \sin^2 	heta & 4 \cos 6	heta \ 0 & 1 & 0 \ -1 & 0 & 1 \end{array} ight| = \left| \begin{array}{ccc} 2 & \sin^2 	heta & 4 \cos 6	heta \ 0 & 1 & 0 \ -1 & 0 & 1 \end{array} ight| = 0$.Expanding along the first column:$2(1 - 0) - 0 + (-1)(0 - 4 \cos 6	heta) = 0$.$2 + 4 \cos 6	heta = 0$.$4 \cos 6	heta = -2 \Rightarrow \cos 6	heta = -\frac{1}{2}$.Since $	heta \in (0, \frac{\pi}{3})$,$6	heta \in (0, 2\pi)$.$\cos 6	heta = -\frac{1}{2} \Rightarrow 6	heta = \frac{2\pi}{3}$ or $6	heta = \frac{4\pi}{3}$.$	heta = \frac{\pi}{9}$ or $	heta = \frac{2\pi}{9}$.Comparing with the options,$	heta = \frac{\pi}{9}$ is the correct value.

$A$ value of $\theta \in (0, \pi /3)$,for which $\left| \begin{array}{ccc} 1 + \cos^2 \theta & \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & 1 + \sin^2 \theta & 4 \cos 6\theta \\ \cos^2 \theta & \sin^2 \theta & 1 + 4 \cos 6\theta \end{array} \right| = 0$,is

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