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 value of \( \theta \) such that the determinant of the given matrix is equal to zero. The matrix is:

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

### Step 1: Calculate the Determinant

To find the determinant, we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where the matrix is represented as:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

For our matrix:

- \( a = \sin 3\theta \)
- \( b = -1 \)
- \( c = 1 \)
- \( d = \cos 2\theta \)
- \( e = 4 \)
- \( f = 3 \)
- \( g = 2 \)
- \( h = 7 \)
- \( i = 7 \)

Now, substituting into the determinant formula:

\[ \text{det} = \sin 3\theta (4 \cdot 7 - 3 \cdot 7) - (-1)(\cos 2\theta \cdot 7 - 3 \cdot 2) + 1(\cos 2\theta \cdot 7 - 4 \cdot 2) \]

Calculating each term:

1. \( 4 \cdot 7 - 3 \cdot 7 = 28 - 21 = 7 \)
2. \( -(-1)(\cos 2\theta \cdot 7 - 6) = \cos 2\theta \cdot 7 - 6 \)
3. \( 1(\cos 2\theta \cdot 7 - 8) = \cos 2\theta \cdot 7 - 8 \)

Putting it all together:

\[ \text{det} = \sin 3\theta \cdot 7 + \cos 2\theta \cdot 7 - 6 + \cos 2\theta \cdot 7 - 8 \]

Combine like terms:

\[ \text{det} = 7\sin 3\theta + 14\cos 2\theta - 14 \]

### Step 2: Set the Determinant to Zero

Now, we set the determinant equal to zero:

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

Dividing through by 7 gives:

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

### Step 3: Rearranging the Equation

Rearranging gives:

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

### Step 4: Use Trigonometric Identities

We know that \( \cos 2\theta = 2\cos^2 \theta - 1 \). Substituting this into the equation gives:

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

This simplifies to:

\[ \sin 3\theta + 4\cos^2 \theta - 2 = 2 \]

Thus,

\[ \sin 3\theta + 4\cos^2 \theta = 4 \]

### Step 5: Solve for \( \theta \)

Since \( \sin 3\theta \) can only take values between -1 and 1, and \( 4\cos^2 \theta \) can take values from 0 to 4, the equation \( \sin 3\theta + 4\cos^2 \theta = 4 \) implies:

\[ \sin 3\theta = 4 - 4\cos^2 \theta \]

The maximum value of \( \sin 3\theta \) is 1, so we need \( 4 - 4\cos^2 \theta \leq 1 \):

\[ 4\cos^2 \theta \geq 3 \implies \cos^2 \theta \geq \frac{3}{4} \]

Thus,

\[ \cos \theta \geq \frac{\sqrt{3}}{2} \text{ or } \cos \theta \leq -\frac{\sqrt{3}}{2} \]

This gives us:

\[ \theta = n\pi \pm \frac{\pi}{6} \]

### Final Answer

The values of \( \theta \) are:

\[ \theta = n\pi + \frac{\pi}{6} \quad \text{or} \quad \theta = n\pi - \frac{\pi}{6} \]