The sum of all possible values of theta in [0 | vedclass

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(D) For a system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.$\begin{vmatrix} \cos 3	h

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$4\pi$

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(D) For a system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.$\begin{vmatrix} \cos 3	heta & -8 & -12 \ \cos 2	heta & 3 & 3 \ 1 & 1 & 3 \end{vmatrix} = 0$Expanding along the first row:$\cos 3	heta(9 - 3) + 8(3\cos 2	heta - 3) - 12(\cos 2	heta - 3) = 0$$6\cos 3	heta + 24\cos 2	heta - 24 - 12\cos 2	heta + 36 = 0$$6\cos 3	heta + 12\cos 2	heta + 12 = 0$Dividing by $6$: $\cos 3	heta + 2\cos 2	heta + 2 = 0$Using trigonometric identities $\cos 3	heta = 4\cos^3	heta - 3\cos	heta$ and $\cos 2	heta = 2\cos^2	heta - 1$:$(4\cos^3	heta - 3\cos	heta) + 2(2\cos^2	heta - 1) + 2 = 0$$4\cos^3	heta + 4\cos^2	heta - 3\cos	heta = 0$$\cos	heta(4\cos^2	heta + 4\cos	heta - 3) = 0$$\cos	heta(2\cos	heta - 1)(2\cos	heta + 3) = 0$Since $\cos	heta$ cannot be $-3/2$, we have $\cos	heta = 0$ or $\cos	heta = 1/2$.For $	heta \in [0, 2\pi]$, $\cos	heta = 0 \Rightarrow 	heta = \pi/2, 3\pi/2$.For $\cos	heta = 1/2 \Rightarrow 	heta = \pi/3, 5\pi/3$.The sum of all values is $\pi/2 + 3\pi/2 + \pi/3 + 5\pi/3 = 2\pi + 2\pi = 4\pi$.

The sum of all possible values of $\theta \in [0, 2\pi]$, for which the system of equations : $x \cos 3\theta - 8y - 12z = 0, x \cos 2\theta + 3y + 3z = 0, x + y + 3z = 0$ has a non-trivial solution, is equal to :

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