Let theta in left(0, frac{pi}{2}right). If th | vedclass

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(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.Let $D = \begin

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$\frac{7 \pi}{18}$

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(B) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.Let $D = \begin{vmatrix} 1+\cos^2 	heta & \sin^2 	heta & 4\sin 3	heta \ \cos^2 	heta & 1+\sin^2 	heta & 4\sin 3	heta \ \cos^2 	heta & \sin^2 	heta & 1+4\sin 3	heta \end{vmatrix} = 0$.Applying $C_1 ightarrow C_1 + C_2$,we get:$D = \begin{vmatrix} 1+\cos^2 	heta + \sin^2 	heta & \sin^2 	heta & 4\sin 3	heta \ \cos^2 	heta + 1+\sin^2 	heta & 1+\sin^2 	heta & 4\sin 3	heta \ \cos^2 	heta + \sin^2 	heta & \sin^2 	heta & 1+4\sin 3	heta \end{vmatrix} = \begin{vmatrix} 2 & \sin^2 	heta & 4\sin 3	heta \ 2 & 1+\sin^2 	heta & 4\sin 3	heta \ 1 & \sin^2 	heta & 1+4\sin 3	heta \end{vmatrix} = 0$.Applying $R_1 ightarrow R_1 - R_2$ and $R_2 ightarrow R_2 - R_3$:$D = \begin{vmatrix} 0 & -1 & 0 \ 1 & 1 & -1 \ 1 & \sin^2 	heta & 1+4\sin 3	heta \end{vmatrix} = 0$.Expanding along the first row:$-(-1) \begin{vmatrix} 1 & -1 \ 1 & 1+4\sin 3	heta \end{vmatrix} = 0 \implies 1(1+4\sin 3	heta) - (-1)(1) = 0$.$1 + 4\sin 3	heta + 1 = 0 \implies 4\sin 3	heta = -2 \implies \sin 3	heta = -\frac{1}{2}$.Since $	heta \in \left(0, \frac{\pi}{2}ight)$,$3	heta \in (0, \frac{3\pi}{2})$.For $\sin 3	heta = -\frac{1}{2}$,$3	heta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.$	heta = \frac{7\pi}{18}$.

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