Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 3 Question 3 of 472 Medium MCQ
The sum of all possible values of θ ∈ [ 0 , 2 π ] \theta \in [0,2\pi ] θ ∈ [ 0 , 2 π ] , for which the system of equations :
x cos 3 θ − 8 y − 12 z = 0 x cos 2 θ + 3 y + 3 z = 0 x + y + 3 z = 0 \begin{gathered} x\cos 3\theta -8y-12z=0 \\ x\cos 2\theta +3y+3z=0 \\ x+y+3z=0 \end{gathered} x cos 3 θ − 8 y − 12 z = 0 x cos 2 θ + 3 y + 3 z = 0 x + y + 3 z = 0
has a non-trivial solution, is equal to :
Solutions · 1 Post your solution Official solution
∣ cos 3 θ − 8 − 12 cos 2 θ 3 3 1 1 3 ∣ = 0 ∣ cos 3 θ − 8 − 4 cos 2 θ 3 1 1 1 1 ∣ = 0 \begin{gathered} |\begin{array}{|c|}\cos 3\theta \quad -8 \quad -12 \\ \cos 2\theta \quad 3 \quad 3 \\ 1 \quad 1 \quad 3\end{array}|=0 \\ |\begin{array}{|c|}\cos 3\theta \quad -8 \quad -4 \\ \cos 2\theta \quad 3 \quad 1 \\ 1 \quad 1 \quad 1\end{array}|=0 \end{gathered} ∣ cos 3 θ − 8 − 12 cos 2 θ 3 3 1 1 3 ∣ = 0 ∣ cos 3 θ − 8 − 4 cos 2 θ 3 1 1 1 1 ∣ = 0
( C 1 → C 1 − C 2 ) C 2 → C 2 − C 3 ∣ cos 3 θ + 8 − 4 − 3 cos 2 θ − 3 2 1 0 0 1 ∣ = 0 \begin{gathered} ({C}_{1}\to {C}_{1}-{C}_{2})\quad {C}_{2}\to {C}_{2}-{C}_{3} \\ |\begin{array}{|c|}\cos 3\theta +8 \quad -4 \quad -3 \\ \cos 2\theta -3 \quad 2 \quad 1 \\ 0 \quad 0 \quad 1\end{array}|=0 \end{gathered} ( C 1 → C 1 − C 2 ) C 2 → C 2 − C 3 ∣ cos 3 θ + 8 − 4 − 3 cos 2 θ − 3 2 1 0 0 1 ∣ = 0
2 cos 3 θ + 16 + 4 cos 2 θ − 12 = 0 ( 4 cos 3 θ − 3 cos θ ) + 2 ( 2 cos 2 θ − 1 ) + 2 = 0 4 cos 3 θ + 4 cos 2 θ − 3 cos θ = 0 cos θ ( 4 cos 2 θ + 4 cos θ − 3 ) = 0 cos θ ( 2 cos θ + 3 ) ( 2 cos θ − 1 ) = 0 cos θ = 0 , 1 2 , − 3 2 θ = π 2 , 3 π 2 , π 3 , 5 π 3 = 4 π \begin{gathered} 2\cos 3\theta +16+4\cos 2\theta -12=0 \\ (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 +3)(2\cos \theta -1)=0 \\ \cos \theta =0,\frac{1}{2},-\frac{3}{2} \\ \theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{5\pi }{3} \\ =4\pi \end{gathered} 2 cos 3 θ + 16 + 4 cos 2 θ − 12 = 0 ( 4 cos 3 θ − 3 cos θ ) + 2 ( 2 cos 2 θ − 1 ) + 2 = 0 4 cos 3 θ + 4 cos 2 θ − 3 cos θ = 0 cos θ ( 4 cos 2 θ + 4 cos θ − 3 ) = 0 cos θ ( 2 cos θ + 3 ) ( 2 cos θ − 1 ) = 0 cos θ = 0 , 2 1 , − 2 3 θ = 2 π , 2 3 π , 3 π , 3 5 π = 4 π Answer the question to unlock