The sum of all possible values of θ ∈ [0, 2π] , for which the...

For the given system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

$\begin{vmatrix} \cos 3\theta & -8 & -12 \\ \cos 2\theta & 3 & 3 \\ 1 & 1 & 3 \end{vmatrix} = 0$

Expanding the determinant along the first row:

$\cos 3\theta (9 - 3) - (-8)(3\cos 2\theta - 3) + (-12)(\cos 2\theta - 3) = 0$

$6\cos 3\theta + 24\cos 2\theta - 24 - 12\cos 2\theta + 36 = 0$

$6\cos 3\theta + 12\cos 2\theta + 12 = 0$

Dividing by $6$, we get:

$\cos 3\theta + 2\cos 2\theta + 2 = 0$

Using the multiple angle formulas $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$ and $\cos 2\theta = 2\cos^2 \theta - 1$, we substitute these into the equation:

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

$4\cos^3 \theta + 4\cos^2 \theta - 3\cos \theta = 0$

$\cos \theta (4\cos^2 \theta + 4\cos \theta - 3) = 0$

$\cos \theta (2\cos \theta - 1)(2\cos \theta + 3) = 0$

This gives the possible values for $\cos \theta$:

$\cos \theta = 0 \implies \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$

$\cos \theta = \dfrac{1}{2} \implies \theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$

$\cos \theta = -\dfrac{3}{2}$ (Not possible since $\cos \theta \in [-1, 1]$)

The possible values of $\theta \in [0, 2\pi]$ are $\dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{\pi}{3}, \dfrac{5\pi}{3}$.

Sum of all possible values of $\theta$:

$\sum \theta = \left(\dfrac{\pi}{2} + \dfrac{3\pi}{2}\right) + \left(\dfrac{\pi}{3} + \dfrac{5\pi}{3}\right) = 2\pi + 2\pi = 4\pi$

Answer: $4\pi$