The number of values of theta in (0, pi) for | vedclass

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(B) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.The coefficient matrix is:$

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$2$

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(B) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.The coefficient matrix is:$A = \begin{bmatrix} 1 & 3 & 7 \ -1 & 4 & 7 \ \sin 3	heta & \cos 2	heta & 2 \end{bmatrix}$Setting the determinant to zero:$\begin{vmatrix} 1 & 3 & 7 \ -1 & 4 & 7 \ \sin 3	heta & \cos 2	heta & 2 \end{vmatrix} = 0$Expanding along the first row:$1(8 - 7\cos 2	heta) - 3(-2 - 7\sin 3	heta) + 7(-\cos 2	heta - 4\sin 3	heta) = 0$$8 - 7\cos 2	heta + 6 + 21\sin 3	heta - 7\cos 2	heta - 28\sin 3	heta = 0$$14 - 14\cos 2	heta - 7\sin 3	heta = 0$$2 - 2\cos 2	heta - \sin 3	heta = 0$Using $\cos 2	heta = 1 - 2\sin^2 	heta$ and $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$:$2 - 2(1 - 2\sin^2 	heta) - (3\sin 	heta - 4\sin^3 	heta) = 0$$2 - 2 + 4\sin^2 	heta - 3\sin 	heta + 4\sin^3 	heta = 0$$4\sin^3 	heta + 4\sin^2 	heta - 3\sin 	heta = 0$$\sin 	heta (4\sin^2 	heta + 4\sin 	heta - 3) = 0$$\sin 	heta (2\sin 	heta - 1)(2\sin 	heta + 3) = 0$Since $	heta \in (0, \pi)$,$\sin 	heta > 0$. Thus,$\sin 	heta = 1/2$ is the only valid solution.$\sin 	heta = 1/2 \implies 	heta = \pi/6, 5\pi/6$.There are $2$ such values.

The number of values of $\theta \in (0, \pi)$ for which the system of linear equations $x + 3y + 7z = 0$,$-x + 4y + 7z = 0$,and $(\sin 3\theta)x + (\cos 2\theta)y + 2z = 0$ has a non-trivial solution is:

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