The system has non trivial solution if `|{:("sin"3theta,-1,1),("cos"2theta,4,3),(2,7,7):}|` =0 then `theta =`

Answer

To solve the problem, we need to find the values of \( \theta \) for which the determinant

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

is equal to zero.

### Step 1: Expand the Determinant
We will expand the determinant using the first row.

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

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 4 & 3 \\ 7 & 7 \end{vmatrix} = 4 \cdot 7 - 3 \cdot 7 = 28 - 21 = 7 \)
2. \( \begin{vmatrix} \cos(2\theta) & 3 \\ 2 & 7 \end{vmatrix} = \cos(2\theta) \cdot 7 - 3 \cdot 2 = 7\cos(2\theta) - 6 \)
3. \( \begin{vmatrix} \cos(2\theta) & 4 \\ 2 & 7 \end{vmatrix} = \cos(2\theta) \cdot 7 - 4 \cdot 2 = 7\cos(2\theta) - 8 \)

Substituting these back into the determinant:

\[ D = \sin(3\theta) \cdot 7 + (7\cos(2\theta) - 6) + (7\cos(2\theta) - 8) \]

### Step 2: Simplify the Expression
Now, we simplify the expression:

\[ D = 7\sin(3\theta) + 7\cos(2\theta) - 6 + 7\cos(2\theta) - 8 \]
\[ D = 7\sin(3\theta) + 14\cos(2\theta) - 14 \]

### Step 3: Set the Determinant to Zero
For the system to have a non-trivial solution, we set the determinant equal to zero:

\[ 7\sin(3\theta) + 14\cos(2\theta) - 14 = 0 \]

Dividing the entire equation by 7 gives:

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

### Step 4: Rearranging the Equation
Rearranging gives us:

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

### Step 5: Use Trigonometric Identities
Using the trigonometric identity for \( \cos(2\theta) \):

\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]

Substituting this into the equation:

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

This simplifies to:

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

### Step 6: Factor the Equation
Factoring gives us:

\[ \sin(3\theta) = 4\sin^2(\theta) \]

### Step 7: Use the Identity for \( \sin(3\theta) \)
Using the identity \( \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) \):

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

Rearranging gives:

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

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

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

### Step 9: Solve for \( \sin(\theta) \)
Setting \( \sin(\theta) = 0 \) gives:

\[ \theta = n\pi \]

For the quadratic \( 4\sin^2(\theta) - 4\sin(\theta) - 3 = 0 \), we can use the quadratic formula:

\[ \sin(\theta) = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} \]
\[ = \frac{4 \pm \sqrt{16 + 48}}{8} \]
\[ = \frac{4 \pm \sqrt{64}}{8} \]
\[ = \frac{4 \pm 8}{8} \]

This gives:

1. \( \sin(\theta) = \frac{12}{8} = \frac{3}{2} \) (not possible since \( \sin(\theta) \) must be in \([-1, 1]\))
2. \( \sin(\theta) = \frac{-4}{8} = -\frac{1}{2} \)

### Step 10: Find \( \theta \) Values
From \( \sin(\theta) = -\frac{1}{2} \):

\[ \theta = \frac{7\pi}{6} + 2n\pi \quad \text{and} \quad \theta = \frac{11\pi}{6} + 2n\pi \]

### Final Answer
Thus, the values of \( \theta \) are:

\[ \theta = n\pi \quad \text{and} \quad \theta = \frac{7\pi}{6} + 2n\pi, \quad \frac{11\pi}{6} + 2n\pi \]