Key Concept: Non-Trivial Solutions for Homogeneous Linear Systems

For a system of homogeneous linear equations, $AX = 0$, where $A$ is the coefficient matrix, $X$ is the variable matrix, and $0$ is the zero matrix, a non-trivial solution exists if and only if the determinant of the coefficient matrix is zero, i.e., $\det(A) = 0$. A non-trivial solution means at least one variable is non-zero.

Step 1: Formulate the Coefficient Matrix and its Determinant

The given system of linear equations is:

1. $x + 3y + 7z = 0$
2. $-x + 4y + 7z = 0$
3. $(\sin 3\theta)x + (\cos 2\theta)y + 2z = 0$

The coefficient matrix $A$ for this system is: $A = \begin{pmatrix} 1 & 3 & 7 \\ -1 & 4 & 7 \\ \sin 3\theta & \cos 2\theta & 2 \end{pmatrix}$

For a non-trivial solution to exist, we must have $\det(A) = 0$. $\det(A) = \left| \begin{matrix} 1 & 3 & 7 \\ -1 & 4 & 7 \\ \sin 3\theta & \cos 2\theta & 2 \end{matrix} \right| = 0$

Step 2: Evaluate the Determinant

We expand the determinant along the first row for clarity. $1 \cdot \left| \begin{matrix} 4 & 7 \\ \cos 2\theta & 2 \end{matrix} \right| - 3 \cdot \left| \begin{matrix} -1 & 7 \\ \sin 3\theta & 2 \end{matrix} \right| + 7 \cdot \left| \begin{matrix} -1 & 4 \\ \sin 3\theta & \cos 2\theta \end{matrix} \right| = 0$

Now, we calculate the $2 \times 2$ determinants: $1 \cdot (4 \cdot 2 - 7 \cdot \cos 2\theta) - 3 \cdot ((-1) \cdot 2 - 7 \cdot \sin 3\theta) + 7 \cdot ((-1) \cdot \cos 2\theta - 4 \cdot \sin 3\theta) = 0$ $1 \cdot (8 - 7 \cos 2\theta) - 3 \cdot (-2 - 7 \sin 3\theta) + 7 \cdot (-\cos 2\theta - 4 \sin 3\theta) = 0$

Distribute the coefficients: $8 - 7 \cos 2\theta + 6 + 21 \sin 3\theta - 7 \cos 2\theta - 28 \sin 3\theta = 0$

Combine like terms: $(8 + 6) + (-7 \cos 2\theta - 7 \cos 2\theta) + (21 \sin 3\theta - 28 \sin 3\theta) = 0$ $14 - 14 \cos 2\theta - 7 \sin 3\theta = 0$

Divide the entire equation by 7 to simplify: $2 - 2 \cos 2\theta - \sin 3\theta = 0$

Step 3: Apply Trigonometric Identities

Our goal is to express the equation in terms of a single trigonometric function, preferably $\sin \theta$, as the interval is $(0, \pi)$, where $\sin \theta \ge 0$. We use the following standard trigonometric identities:

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

Substitute these identities into the simplified equation $2 - 2 \cos 2\theta - \sin 3\theta = 0$: $2 - 2(1 - 2 \sin^2 \theta) - (3 \sin \theta - 4 \sin^3 \theta) = 0$

Expand and simplify: $2 - 2 + 4 \sin^2 \theta - 3 \sin \theta + 4 \sin^3 \theta = 0$ $4 \sin^3 \theta + 4 \sin^2 \theta - 3 \sin \theta = 0$

Step 4: Solve the Trigonometric Equation

Now we have a cubic equation in terms of $\sin \theta$. We can factor out $\sin \theta$: $\sin \theta (4 \sin^2 \theta + 4 \sin \theta - 3) = 0$

This equation gives us two possibilities:

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

Let's solve the quadratic equation $4 \sin^2 \theta + 4 \sin \theta - 3 = 0$. Let $x = \sin \theta$. $4x^2 + 4x - 3 = 0$ We can solve this quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-4 \pm \sqrt{4^2 - 4(4)(-3)}}{2(4)}$ $x = \frac{-4 \pm \sqrt{16 + 48}}{8}$ $x = \frac{-4 \pm \sqrt{64}}{8}$ $x = \frac{-4 \pm 8}{8}$

This yields two possible values for $x = \sin \theta$:

• $\sin \theta = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2}$
• $\sin \theta = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2}$

Step 5: Find Values of $\theta$ in the Interval $(0, \pi)$

We need to find the number of values of $\theta$ in the interval $(0, \pi)$ that satisfy any of these conditions:

• Case 1: $\sin \theta = 0$ In the interval $(0, \pi)$, $\sin \theta = 0$ has no solutions. The solutions for $\sin \theta = 0$ are $\theta = n\pi$ for integer $n$. The values $0$ and $\pi$ are excluded from the open interval $(0, \pi)$.

• Case 2: $\sin \theta = \frac{1}{2}$ In the interval $(0, \pi)$, $\sin \theta = \frac{1}{2}$ has two solutions:

  • $\theta = \frac{\pi}{6}$ (in the first quadrant)
  • $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ (in the second quadrant) Both of these values are strictly between $0$ and $\pi$.

• Case 3: $\sin \theta = -\frac{3}{2}$ The sine function has a range of $[-1, 1]$. Since $-\frac{3}{2}$ is less than $-1$, there are no real values of $\theta$ for which $\sin \theta = -\frac{3}{2}$.

Step 6: Count the Solutions

Combining all valid solutions, we have two values of $\theta$ in the interval $(0, \pi)$: $\frac{\pi}{6}$ and $\frac{5\pi}{6}$.

Therefore, the number of values of $\theta$ for which the system has a non-trivial solution is 2.

Tips and Common Mistakes:

• Remember the condition: Always start by recalling that a homogeneous system has non-trivial solutions if and only if the determinant of the coefficient matrix is zero.
• Determinant Calculation: Be meticulous with signs and arithmetic when calculating the determinant. A small error here can propagate through the entire problem.
• Trigonometric Identities: Memorize or be able to quickly derive common identities like $\sin 2\theta$, $\cos 2\theta$, $\sin 3\theta$, $\cos 3\theta$. Choosing the right identity (e.g., $\cos 2\theta = 1 - 2\sin^2\theta$ when converting to $\sin\theta$) is crucial.
• Domain and Range: Always consider the given interval for $\theta$ (here, $(0, \pi)$) and the range of trigonometric functions (e.g., $\sin \theta \in [-1, 1]$). This helps in eliminating extraneous solutions.
• Factoring: Don't forget to factor out common terms (like $\sin \theta$ in this case) before solving quadratic or higher-order equations.

Summary:

We determined the condition for non-trivial solutions by setting the determinant of the coefficient matrix to zero. This led to a trigonometric equation involving $\sin 3\theta$ and $\cos 2\theta$. By applying appropriate trigonometric identities to express everything in terms of $\sin \theta$, we converted the equation into a cubic polynomial in $\sin \theta$. Factoring and solving this polynomial yielded possible values for $\sin \theta$. Finally, by carefully checking these values within the given interval $(0, \pi)$ for $\theta$, we found exactly two solutions.

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