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(B) For the system of linear equations to have no solution,the determinant $D$ of the coefficient matrix must be zero,and the system must be inconsist

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$7$

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(B) For the system of linear equations to have no solution,the determinant $D$ of the coefficient matrix must be zero,and the system must be inconsistent.First,calculate the determinant $D$:$D = \begin{vmatrix} 3\sin 3	heta & -1 & 1 \ 3\cos 2	heta & 4 & 3 \ 6 & 7 & 7 \end{vmatrix} = 0$$D = 3\sin 3	heta(28 - 21) + 1(21\cos 2	heta - 18) + 1(21\cos 2	heta - 24) = 0$$D = 3\sin 3	heta(7) + 21\cos 2	heta - 18 + 21\cos 2	heta - 24 = 0$$21\sin 3	heta + 42\cos 2	heta - 42 = 0$Dividing by $21$,we get $\sin 3	heta + 2\cos 2	heta - 2 = 0$.Using $\cos 2	heta = 1 - 2\sin^2 	heta$ and $\sin 3	heta = 3\sin 	heta - 4\sin^3 	heta$:$(3\sin 	heta - 4\sin^3 	heta) + 2(1 - 2\sin^2 	heta) - 2 = 0$$3\sin 	heta - 4\sin^3 	heta + 2 - 4\sin^2 	heta - 2 = 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$This gives $\sin 	heta = 0$,$\sin 	heta = 1/2$,or $\sin 	heta = -3/2$ (impossible).For $\sin 	heta = 0$ in $(0, 4\pi)$,$	heta = \pi, 2\pi, 3\pi$.For $\sin 	heta = 1/2$ in $(0, 4\pi)$,$	heta = \pi/6, 5\pi/6, 13\pi/6, 17\pi/6$.Checking consistency,all these values lead to no solution. Thus,there are $3 + 4 = 7$ values.

The number of $\theta \in (0, 4\pi)$ for which the system of linear equations $3(\sin 3\theta)x - y + z = 2$,$3(\cos 2\theta)x + 4y + 3z = 3$,and $6x + 7y + 7z = 9$ has no solution is:

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