1. Condition for no solution

For the system

$$\begin{aligned} 3(\sin 3\theta)x-y+z&=2 \\ 3(\cos 2\theta)x+4y+3z&=3 \\ 6x+7y+7z&=9 \end{aligned}$$

to have no solution, we need:

• the coefficient matrix to be singular, and
• the augmented system to be inconsistent.

So let the coefficient matrix be

$$A=\begin{pmatrix} 3\sin 3\theta & -1 & 1\\ 3\cos 2\theta & 4 & 3\\ 6 & 7 & 7 \end{pmatrix}.$$

We first find when $\det A=0$.

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2. Compute the determinant

$$\det A= \begin{vmatrix} 3\sin 3\theta & -1 & 1\\ 3\cos 2\theta & 4 & 3\\ 6 & 7 & 7 \end{vmatrix}$$

Expanding along the first row:

$$\det A=3\sin 3\theta \begin{vmatrix} 4 & 3\\ 7 & 7 \end{vmatrix} -(-1) \begin{vmatrix} 3\cos 2\theta & 3\\ 6 & 7 \end{vmatrix} +1 \begin{vmatrix} 3\cos 2\theta & 4\\ 6 & 7 \end{vmatrix}$$

Now,

$$\begin{vmatrix} 4 & 3\\ 7 & 7 \end{vmatrix}=28-21=7,$$ $$\begin{vmatrix} 3\cos 2\theta & 3\\ 6 & 7 \end{vmatrix}=21\cos 2\theta-18,$$ $$\begin{vmatrix} 3\cos 2\theta & 4\\ 6 & 7 \end{vmatrix}=21\cos 2\theta-24.$$

Hence

$$\det A=21\sin 3\theta+(21\cos 2\theta-18)+(21\cos 2\theta-24).$$

So

$$\det A=21\sin 3\theta+42\cos 2\theta-42.$$

Thus

$$\det A=21\big(\sin 3\theta+2\cos 2\theta-2\big).$$

Therefore singularity requires

$$\sin 3\theta+2\cos 2\theta-2=0.$$

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3. Simplify the trigonometric condition

Use

$$\sin 3\theta=\sin\theta(1+2\cos 2\theta).$$

Then

$$\sin 3\theta+2\cos 2\theta-2=0.$$

A more useful identity is obtained by rewriting in terms of $\sin\theta$:

$$\sin 3\theta=3\sin\theta-4\sin^3\theta,$$ $$\cos 2\theta=1-2\sin^2\theta.$$

So

$$3\sin\theta-4\sin^3\theta+2(1-2\sin^2\theta)-2=0.$$

This gives

$$3\sin\theta-4\sin^3\theta-4\sin^2\theta=0.$$

Factor:

$$\sin\theta\,(3-4\sin^2\theta-4\sin\theta)=0.$$

So either

$$\sin\theta=0$$

or

$$4\sin^2\theta+4\sin\theta-3=0.$$

Let $s=\sin\theta$. Then

$$4s^2+4s-3=0$$ $$(2s-1)(2s+3)=0.$$

Thus

$$s=\frac12 \quad \text{or} \quad s=-\frac32.$$

The second is impossible. Hence

$$\sin\theta=0 \quad \text{or} \quad \sin\theta=\frac12.$$

In $(0,4\pi)$, these give:

• $\sin\theta=0 \Rightarrow \theta=\pi,2\pi,3\pi$,
• $\sin\theta=\frac12 \Rightarrow \theta=\frac\pi6,\frac{5\pi}6,\frac{13\pi}6,\frac{17\pi}6$.

So there are 7 values for which $\det A=0$.

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4. Check whether these singular cases are inconsistent

We must ensure the system has no solution, not infinitely many solutions.

Observe the third equation minus the second equation:

$$(6-3\cos2\theta)x+(7-4)y+(7-3)z=9-3,$$

so

$$(6-3\cos2\theta)x+3y+4z=6.$$

This does not immediately show consistency.

Instead, when $\det A=0$, check whether the first row becomes a linear combination of the other two in the coefficient matrix and whether constants match.

Let us test the singular values.

Case 1: $\sin\theta=0$

Then $\theta=\pi,2\pi,3\pi$.

Now,

$$\sin 3\theta=0, \qquad \cos 2\theta=1.$$

The system becomes

$$-y+z=2,$$ $$3x+4y+3z=3,$$ $$6x+7y+7z=9.$$

From the first equation,

$$z=y+2.$$

Substitute into the second:

$$3x+4y+3(y+2)=3 \Rightarrow 3x+7y=-3.$$

Substitute into the third:

$$6x+7y+7(y+2)=9 \Rightarrow 6x+14y=-5.$$

But doubling $3x+7y=-3$ gives

$$6x+14y=-6,$$

which contradicts $6x+14y=-5$.

So these 3 values give no solution.

Case 2: $\sin\theta=\frac12$

Then $\theta=\frac\pi6,\frac{5\pi}6,\frac{13\pi}6,\frac{17\pi}6$.

For all these,

$$\cos 2\theta=\frac12,$$

and since they also satisfy the determinant condition,

$$\sin 3\theta=1.$$

Thus the system becomes

$$3x-y+z=2,$$ $$\frac32 x+4y+3z=3,$$ $$6x+7y+7z=9.$$

Multiply the second equation by $2$:

$$3x+8y+6z=6.$$

Subtract the first equation:

$$9y+5z=4.$$

Also from third minus twice first:

$$(6x+7y+7z)-(6x-2y+2z)=9-4,$$

so

$$9y+5z=5.$$

This is impossible.

Hence these 4 values also give no solution.

Therefore all 7 singular values correspond to inconsistency.

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5. Total number of $\theta$

$$3+4=7.$$

So the required number is

$$\boxed{7}.$$

Thus the correct option is B.

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6. Comparison with stored answer

Stored correct answer: B

Derived answer: B

So they agree.