To find the sum of all possible values of θ ∈ [ 0 , 2 π ] \theta \in [0, 2\pi] θ ∈ [ 0 , 2 π ]
The given homogeneous system of linear equations is:
x cos 3 θ − 8 y − 12 z = 0 x cos 2 θ + 3 y + 3 z = 0 x + y + 3 z = 0 \begin{aligned}
x\cos 3\theta - 8y - 12z &= 0 \\
x\cos 2\theta + 3y + 3z &= 0 \\
x + y + 3z &= 0
\end{aligned} x cos 3 θ − 8 y − 12 z x cos 2 θ + 3 y + 3 z x + y + 3 z = 0 = 0 = 0 For a non-trivial solution, the coefficient determinant D D D
D = ∣ cos 3 θ − 8 − 12 cos 2 θ 3 3 1 1 3 ∣ = 0 D = \begin{vmatrix}
\cos 3\theta & -8 & -12 \\
\cos 2\theta & 3 & 3 \\
1 & 1 & 3
\end{vmatrix} = 0 D = cos 3 θ cos 2 θ 1 − 8 3 1 − 12 3 3 = 0 Expanding the determinant along the first row:
D = cos 3 θ ∣ 3 3 1 3 ∣ − ( − 8 ) ∣ cos 2 θ 3 1 3 ∣ + ( − 12 ) ∣ cos 2 θ 3 1 1 ∣ = 0 D = \cos 3\theta \begin{vmatrix} 3 & 3 \\ 1 & 3 \end{vmatrix} - (-8) \begin{vmatrix} \cos 2\theta & 3 \\ 1 & 3 \end{vmatrix} + (-12) \begin{vmatrix} \cos 2\theta & 3 \\ 1 & 1 \end{vmatrix} = 0 D = cos 3 θ 3 1 3 3 − ( − 8 ) cos 2 θ 1 3 3 + ( − 12 ) cos 2 θ 1 3 1 = 0 Calculating the 2 × 2 2 \times 2 2 × 2
∣ 3 3 1 3 ∣ = 3 ( 3 ) − 3 ( 1 ) = 9 − 3 = 6 ∣ cos 2 θ 3 1 3 ∣ = 3 cos 2 θ − 3 ∣ cos 2 θ 3 1 1 ∣ = cos 2 θ − 3 \begin{aligned}
\begin{vmatrix} 3 & 3 \\ 1 & 3 \end{vmatrix} &= 3(3) - 3(1) = 9 - 3 = 6 \\
\begin{vmatrix} \cos 2\theta & 3 \\ 1 & 3 \end{vmatrix} &= 3\cos 2\theta - 3 \\
\begin{vmatrix} \cos 2\theta & 3 \\ 1 & 1 \end{vmatrix} &= \cos 2\theta - 3
\end{aligned} 3 1 3 3 cos 2 θ 1 3 3 cos 2 θ 1 3 1 = 3 ( 3 ) − 3 ( 1 ) = 9 − 3 = 6 = 3 cos 2 θ − 3 = cos 2 θ − 3 Substitute these back into D D D
D = 6 cos 3 θ + 8 ( 3 cos 2 θ − 3 ) − 12 ( cos 2 θ − 3 ) = 0 D = 6\cos 3\theta + 8(3\cos 2\theta - 3) - 12(\cos 2\theta - 3) = 0 D = 6 cos 3 θ + 8 ( 3 cos 2 θ − 3 ) − 12 ( cos 2 θ − 3 ) = 0 6 cos 3 θ + 24 cos 2 θ − 24 − 12 cos 2 θ + 36 = 0 6\cos 3\theta + 24\cos 2\theta - 24 - 12\cos 2\theta + 36 = 0 6 cos 3 θ + 24 cos 2 θ − 24 − 12 cos 2 θ + 36 = 0 6 cos 3 θ + 12 cos 2 θ + 12 = 0 6\cos 3\theta + 12\cos 2\theta + 12 = 0 6 cos 3 θ + 12 cos 2 θ + 12 = 0 Dividing the entire equation by 6 6 6
cos 3 θ + 2 cos 2 θ + 2 = 0 \cos 3\theta + 2\cos 2\theta + 2 = 0 cos 3 θ + 2 cos 2 θ + 2 = 0 Using the trigonometric identities cos 3 θ = 4 cos 3 θ − 3 cos θ \cos 3\theta = 4\cos^3 \theta - 3\cos \theta cos 3 θ = 4 cos 3 θ − 3 cos θ cos 2 θ = 2 cos 2 θ − 1 \cos 2\theta = 2\cos^2 \theta - 1 cos 2 θ = 2 cos 2 θ − 1 u = cos θ u = \cos \theta u = cos θ
( 4 u 3 − 3 u ) + 2 ( 2 u 2 − 1 ) + 2 = 0 (4u^3 - 3u) + 2(2u^2 - 1) + 2 = 0 ( 4 u 3 − 3 u ) + 2 ( 2 u 2 − 1 ) + 2 = 0 4 u 3 + 4 u 2 − 3 u = 0 4u^3 + 4u^2 - 3u = 0 4 u 3 + 4 u 2 − 3 u = 0 u ( 4 u 2 + 4 u − 3 ) = 0 u(4u^2 + 4u - 3) = 0 u ( 4 u 2 + 4 u − 3 ) = 0 Factorizing the quadratic expression 4 u 2 + 4 u − 3 4u^2 + 4u - 3 4 u 2 + 4 u − 3
u ( 2 u − 1 ) ( 2 u + 3 ) = 0 u(2u - 1)(2u + 3) = 0 u ( 2 u − 1 ) ( 2 u + 3 ) = 0 This gives three possible cases for u = cos θ u = \cos \theta u = cos θ
u = 0 ⟹ cos θ = 0 u = 0 \implies \cos \theta = 0 u = 0 ⟹ cos θ = 0 u = 1 2 ⟹ cos θ = 1 2 u = \frac{1}{2} \implies \cos \theta = \frac{1}{2} u = 2 1 ⟹ cos θ = 2 1 u = − 3 2 ⟹ cos θ = − 3 2 u = -\frac{3}{2} \implies \cos \theta = -\frac{3}{2} u = − 2 3 ⟹ cos θ = − 2 3 cos θ ∈ [ − 1 , 1 ] \cos \theta \in [-1, 1] cos θ ∈ [ − 1 , 1 ]
Now, we find the values of θ ∈ [ 0 , 2 π ] \theta \in [0, 2\pi] θ ∈ [ 0 , 2 π ]
Thus, the set of all possible values for θ \theta θ [ 0 , 2 π ] [0, 2\pi] [ 0 , 2 π ]
θ ∈ { π 3 , π 2 , 3 π 2 , 5 π 3 } \theta \in \left\{ \frac{\pi}{3}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{3} \right\} θ ∈ { 3 π , 2 π , 2 3 π , 3 5 π } Sum of all these values:
Sum = π 3 + π 2 + 3 π 2 + 5 π 3 = ( π 3 + 5 π 3 ) + ( π 2 + 3 π 2 ) = 2 π + 2 π = 4 π \text{Sum} = \frac{\pi}{3} + \frac{\pi}{2} + \frac{3\pi}{2} + \frac{5\pi}{3} = \left(\frac{\pi}{3} + \frac{5\pi}{3}\right) + \left(\frac{\pi}{2} + \frac{3\pi}{2}\right) = 2\pi + 2\pi = 4\pi Sum = 3 π + 2 π + 2 3 π + 3 5 π = ( 3 π + 3 5 π ) + ( 2 π + 2 3 π ) = 2 π + 2 π = 4 π Hence, the correct option is D .