Question

Sol. [Key Concept: A non-trivial solution needs the coefficient determinant $= 0$; expand it into a cubic in $cos\theta$, solve, and add all $\theta$ in $\lbrack 0,2\pi\rbrack$.]

$$\left| \begin{matrix} \cos{3\theta} & - 8 & - 12 \\ \cos{2\theta} & 3 & 3 \\ 1 & 1 & 3 \end{matrix} \right| = 0$$

$$ {\left| \begin{matrix} \cos{3\theta} & - 8 & - 4 \\ \cos{2\theta} & 3 & 1 \\ 1 & 1 & 1 \end{matrix} \right| = 0}$$

$$ {\left( C_{1} \rightarrow C_{1} - C_{2} \right)\ \&{\ C}_{2} \rightarrow C_{2} - C_{3}}$$

$$ {\left| \begin{matrix} cos3\theta + 8 & - 4 & - 3 \\ cos2\theta - 3 & 2 & 1 \\ 0 & 0 & 1 \end{matrix} \right| = 0}$$

$$ {2cos3\theta + 16 + 4cos2\theta - 12 = 0}$$

$$ {\left( 4\cos^{3}\theta - 3\cos\theta \right) + 2\left( 2\cos^{2}\theta - 1 \right) + 2 = 0}$$

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

$$ {cos\theta\left( 4\cos^{2}\theta + 4cos\theta - 3 \right) = 0}$$

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

$$ cos\theta = 0,\frac{1}{2}, - \frac{3}{2}$$ (rejected)

$$\theta = \frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{3},\frac{5\pi}{3}$$

$$ {Sum = 4\pi }$$