Find the value of `theta` if `|[1,1,sin 3theta] , [-4,3,cos 2theta] , [7, -7, -2]|=0`

Explanation

To find the value of `theta` such that the determinant

\[ \begin{vmatrix} 1 & 1 & \sin(3\theta) \\ -4 & 3 & \cos(2\theta) \\ 7 & -7 & -2 \end{vmatrix} = 0, \]

we will follow these steps:

### Step 1: Apply Column Transformation
We can simplify the determinant by applying the transformation \( C_1 \rightarrow C_1 - C_2 \).

\[ \begin{vmatrix} 1 - 1 & 1 & \sin(3\theta) \\ -4 - 3 & 3 & \cos(2\theta) \\ 7 - (-7) & -7 & -2 \end{vmatrix} = \begin{vmatrix} 0 & 1 & \sin(3\theta) \\ -7 & 3 & \cos(2\theta) \\ 14 & -7 & -2 \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Now we can factor out common terms from the determinant. Notice that the second row can be simplified:

\[ = 7 \cdot \begin{vmatrix} 0 & 1 & \sin(3\theta) \\ -1 & 3 & \cos(2\theta) \\ 2 & -1 & -2 \end{vmatrix} \]

### Step 3: Expand the Determinant
Now we can expand the determinant along the first column:

\[ = 7 \cdot \left( 0 \cdot \begin{vmatrix} 3 & \cos(2\theta) \\ -1 & -2 \end{vmatrix} - 1 \cdot \begin{vmatrix} -1 & \cos(2\theta) \\ 2 & -2 \end{vmatrix} + \sin(3\theta) \cdot \begin{vmatrix} -1 & 3 \\ 2 & -1 \end{vmatrix} \right) \]

Calculating the 2x2 determinants:

\[ = 7 \cdot \left( -\begin{vmatrix} -1 & \cos(2\theta) \\ 2 & -2 \end{vmatrix} + \sin(3\theta) \cdot \begin{vmatrix} -1 & 3 \\ 2 & -1 \end{vmatrix} \right) \]

Calculating these determinants:

1. \(\begin{vmatrix} -1 & \cos(2\theta) \\ 2 & -2 \end{vmatrix} = (-1)(-2) - (2)(\cos(2\theta)) = 2 - 2\cos(2\theta)\)

2. \(\begin{vmatrix} -1 & 3 \\ 2 & -1 \end{vmatrix} = (-1)(-1) - (3)(2) = 1 - 6 = -5\)

### Step 4: Substitute Back
Substituting back into the determinant:

\[ = 7 \cdot \left( - (2 - 2\cos(2\theta)) - 5\sin(3\theta) \right) = 7 \cdot \left( -2 + 2\cos(2\theta) + 5\sin(3\theta) \right) \]

Setting the determinant equal to zero:

\[ -2 + 2\cos(2\theta) + 5\sin(3\theta) = 0 \]

### Step 5: Rearranging the Equation
Rearranging gives:

\[ 2\cos(2\theta) + 5\sin(3\theta) = 2 \]

### Step 6: Use Trigonometric Identities
Using the identities \(\cos(2\theta) = 1 - 2\sin^2(\theta)\) and \(\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)\):

Substituting these identities into the equation:

\[ 2(1 - 2\sin^2(\theta)) + 5(3\sin(\theta) - 4\sin^3(\theta)) = 2 \]

### Step 7: Simplifying the Equation
This simplifies to:

\[ 2 - 4\sin^2(\theta) + 15\sin(\theta) - 20\sin^3(\theta) = 2 \]

This leads to:

\[ -4\sin^2(\theta) + 15\sin(\theta) - 20\sin^3(\theta) = 0 \]

### Step 8: Factor the Equation
Factoring out \(\sin(\theta)\):

\[ \sin(\theta)(-20\sin^2(\theta) - 4\sin(\theta) + 15) = 0 \]

### Step 9: Solve for \(\sin(\theta)\)
This gives us two cases:

1. \(\sin(\theta) = 0\)
2. \(-20\sin^2(\theta) - 4\sin(\theta) + 15 = 0\)

### Step 10: Finding Values of \(\theta\)
For \(\sin(\theta) = 0\), we have:

\[ \theta = n\pi, \quad n \in \mathbb{Z} \]

For the quadratic equation, we can use the quadratic formula to find the roots.

### Final Answer
The values of \(\theta\) that satisfy the original determinant equation are:

\[ \theta = n\pi \quad \text{and the solutions from the quadratic equation.} \]