This problem asks us to determine the number of values of $\theta$ in the interval $(0, 4\pi)$ for which a given system of three linear equations in $x, y, z$ has no solution.

1. Key Concept: Conditions for a System of Linear Equations For a system of linear equations $A\mathbf{x} = \mathbf{b}$, where $A$ is the coefficient matrix, $\mathbf{x}$ is the variable matrix, and $\mathbf{b}$ is the constant matrix:

• Unique Solution: The system has a unique solution if and only if $\det(A) \neq 0$.
• No Solution (Inconsistent System): The system has no solution if and only if $\det(A) = 0$ AND at least one of the determinants $\det(A_x)$, $\det(A_y)$, or $\det(A_z)$ is non-zero. (Here, $A_x, A_y, A_z$ are matrices formed by replacing the respective coefficient columns with the constant column $\mathbf{b}$, as per Cramer's Rule).
• Infinitely Many Solutions: The system has infinitely many solutions if and only if $\det(A) = 0$ AND $\det(A_x) = \det(A_y) = \det(A_z) = 0$.

Alternatively, using rank:

• Unique Solution: $\text{rank}(A) = \text{rank}([A|\mathbf{b}]) = \text{number of variables}$.
• No Solution: $\text{rank}(A) \neq \text{rank}([A|\mathbf{b}])$.
• Infinitely Many Solutions: $\text{rank}(A) = \text{rank}([A|\mathbf{b}]) < \text{number of variables}$.

2. Formulate the Coefficient Matrix and Constant Vector The given system of equations is: $\begin{aligned} &3(\sin 3 \theta) x-y+z=2 \\ &3(\cos 2 \theta) x+4 y+3 z=3 \\ &6 x+7 y+7 z=9 \end{aligned}$ The coefficient matrix $A$ and the constant vector $\mathbf{b}$ are: $A = \begin{pmatrix} 3\sin 3\theta & -1 & 1 \\ 3\cos 2\theta & 4 & 3 \\ 6 & 7 & 7 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ 9 \end{pmatrix}$

3. Calculate the Determinant of the Coefficient Matrix, $\det(A)$ For the system to have no solution, the primary condition is $\det(A) = 0$. Let's compute $\det(A)$: $\det(A) = (3\sin 3\theta) \begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} - (-1) \begin{vmatrix} 3\cos 2\theta & 3 \\ 6 & 7 \end{vmatrix} + (1) \begin{vmatrix} 3\cos 2\theta & 4 \\ 6 & 7 \end{vmatrix}$ $\det(A) = (3\sin 3\theta)(4 \cdot 7 - 3 \cdot 7) + (1)(3\cos 2\theta \cdot 7 - 3 \cdot 6) + (1)(3\cos 2\theta \cdot 7 - 4 \cdot 6)$ $\det(A) = (3\sin 3\theta)(28 - 21) + (21\cos 2\theta - 18) + (21\cos 2\theta - 24)$ $\det(A) = 3\sin 3\theta (7) + 21\cos 2\theta - 18 + 21\cos 2\theta - 24$ $\det(A) = 21\sin 3\theta + 42\cos 2\theta - 42$

4. Set $\det(A) = 0$ and Solve for $\theta$ $21\sin 3\theta + 42\cos 2\theta - 42 = 0$ Divide the entire equation by 21: $\sin 3\theta + 2\cos 2\theta - 2 = 0$ Now, we use trigonometric identities to express $\sin 3\theta$ and $\cos 2\theta$ in terms of $\sin\theta$:

• $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$

• $\cos 2\theta = 1 - 2\sin^2\theta$ Substitute these into the equation: $(3\sin\theta - 4\sin^3\theta) + 2(1 - 2\sin^2\theta) - 2 = 0$ $3\sin\theta - 4\sin^3\theta + 2 - 4\sin^2\theta - 2 = 0$ $3\sin\theta - 4\sin^3\theta - 4\sin^2\theta = 0$ Factor out $\sin\theta$: $\sin\theta (3 - 4\sin^2\theta - 4\sin\theta) = 0$ Rearrange the quadratic term: $\sin\theta (4\sin^2\theta + 4\sin\theta - 3) = 0$ This equation gives two possibilities:

• Case 1: $\sin\theta = 0$ For $\theta \in (0, 4\pi)$, the solutions are $\theta = \pi, 2\pi, 3\pi$. (3 values)

• Case 2: $4\sin^2\theta + 4\sin\theta - 3 = 0$ Let $s = \sin\theta$. This is a quadratic equation: $4s^2 + 4s - 3 = 0$. Using the quadratic formula $s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $s = \frac{-4 \pm \sqrt{4^2 - 4(4)(-3)}}{2(4)} = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm \sqrt{64}}{8} = \frac{-4 \pm 8}{8}$ Two possible values for $s$: $s_1 = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2}$ $s_2 = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2}$ Since $\sin\theta$ must be in the range $[-1, 1]$, $s_2 = -\frac{3}{2}$ is not a valid solution. So, we only consider $\sin\theta = \frac{1}{2}$. For $\theta \in (0, 4\pi)$, the solutions for $\sin\theta = \frac{1}{2}$ are: In $(0, 2\pi)$: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ In $(2\pi, 4\pi)$: $\theta = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$, and $\theta = 2\pi + \frac{5\pi}{6} = \frac{17\pi}{6}$ (4 values)

Combining both cases, the values of $\theta$ for which $\det(A) = 0$ are: $\left\{ \pi, 2\pi, 3\pi, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6} \right\}$ This gives a total of $3 + 4 = 7$ values.

5. Check for "No Solution" Condition For the system to have no solution, we need $\det(A) = 0$ AND at least one of $\det(A_x)$, $\det(A_y)$, or $\det(A_z)$ to be non-zero. Let's compute $\det(A_x)$: $A_x = \begin{pmatrix} 2 & -1 & 1 \\ 3 & 4 & 3 \\ 9 & 7 & 7 \end{pmatrix}$ $\det(A_x) = (2) \begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} - (-1) \begin{vmatrix} 3 & 3 \\ 9 & 7 \end{vmatrix} + (1) \begin{vmatrix} 3 & 4 \\ 9 & 7 \end{vmatrix}$ $\det(A_x) = (2)(4 \cdot 7 - 3 \cdot 7) + (1)(3 \cdot 7 - 3 \cdot 9) + (1)(3 \cdot 7 - 4 \cdot 9)$ $\det(A_x) = (2)(28 - 21) + (21 - 27) + (21 - 36)$ $\det(A_x) = 2(7) + (-6) + (-15)$ $\det(A_x) = 14 - 6 - 15 = 8 - 15 = -7$ Since $\det(A_x) = -7$, which is a non-zero constant, it means that for any value of $\theta$ for which $\det(A)=0$, the condition $\det(A_x) \neq 0$ is always satisfied. Therefore, all 7 values of $\theta$ found in step 4 will result in the system having no solution.

Tip: A common mistake is to only check $\det(A)=0$ and assume all such values lead to no solution. It's crucial to verify that the system doesn't have infinitely many solutions for these values. In this case, $\det(A_x) \neq 0$ simplifies this check significantly.

6. Alternative Verification using Rank (Advanced) The condition $\det(A)=0$ implies that the rows of $A$ are linearly dependent. Let $R_1, R_2, R_3$ be the rows of $A$. We found that $\det(A)=0$ when $\sin 3\theta + 2\cos 2\theta - 2 = 0$. Let's check if $R_3$ can be expressed as a linear combination of $R_1$ and $R_2$, i.e., $R_3 = c_1 R_1 + c_2 R_2$. Comparing the $y$ and $z$ coefficients: $-c_1 + 4c_2 = 7$ $c_1 + 3c_2 = 7$ Adding these two equations gives $7c_2 = 14 \implies c_2 = 2$. Substituting $c_2 = 2$ into $c_1 + 3c_2 = 7$ gives $c_1 + 6 = 7 \implies c_1 = 1$. So, the relation is $R_3 = R_1 + 2R_2$. Now, let's verify this relation for the $x$ coefficients: $6 = (3\sin 3\theta) + 2(3\cos 2\theta) \implies 6 = 3\sin 3\theta + 6\cos 2\theta \implies 2 = \sin 3\theta + 2\cos 2\theta$. This is precisely the equation we solved for $\det(A)=0$. This means that for all values of $\theta$ where $\det(A)=0$, we have $\text{rank}(A)=2$ (since $R_1$ and $R_2$ are clearly linearly independent). Now, we check the augmented matrix $[A|\mathbf{b}]$. For the system to have no solution, $\text{rank}(A) \neq \text{rank}([A|\mathbf{b}])$. The linear relation $R_3 = R_1 + 2R_2$ for the coefficient part must NOT hold for the constant terms. Let's check if $b_3 = c_1 b_1 + c_2 b_2$: $9 = 1 \cdot (2) + 2 \cdot (3)$ $9 = 2 + 6$ $9 = 8$ This statement $9=8$ is false. This means the linear dependency among the rows of $A$ does not extend to the augmented matrix $[A|\mathbf{b}]$. Therefore, $\text{rank}([A|\mathbf{b}]) = 3$. Since $\text{rank}(A) = 2$ and $\text{rank}([A|\mathbf{b}]) = 3$, we have $\text{rank}(A) \neq \text{rank}([A|\mathbf{b}])$, which confirms that the system has no solution for all 7 values of $\theta$ found.

7. Final Count The values of $\theta \in (0, 4\pi)$ for which the system has no solution are: $\pi, 2\pi, 3\pi$ (from $\sin\theta = 0$) $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$ (from $\sin\theta = \frac{1}{2}$) All these 7 values are distinct and lie within the interval $(0, 4\pi)$.

Summary and Key Takeaway: To determine when a system of linear equations has no solution, we first set the determinant of the coefficient matrix to zero. Then, we must verify that for these values, the system does not have infinitely many solutions. This is done by checking if at least one of the determinants $A_x, A_y, A_z$ is non-zero. In this problem, $\det(A_x)$ was found to be a constant non-zero value, which simplified the verification, ensuring that all values of $\theta$ making $\det(A)=0$ lead to no solution.

The total number of values of $\theta$ is $\mathbf{7}$.

The final answer is $\boxed{\text{7}}$.