Mathematics · Determinants

JEE Main 2026 — 6 April, Evening Shift — Question 24

The sum of all possible values of θ∈[0,2π]\theta \in [0, 2\pi ], for which the system of equations: xcos3θ−8y−12z=0,xcos2θ+3y+3z=0,x+y+3z=0x cos3θ - 8y - 12z = 0, x cos2θ + 3y + 3z = 0, x + y + 3z = 0 has a non-trivial solution, is equal to:

  1. Option A:

    π

  2. Option B:

    2π

  3. Option C:

    3π

  4. Option D:

    4π

    Correct

Answer: D

Step-by-step solution

∣cos⁡3θ−8−12cos⁡2θ33113∣=0\left|\begin{array}{ccc}\cos 3 \theta & -8 & -12\\ \cos 2 \theta & 3 & 3\\ 1 & 1 & 3\end{array}\right|=0 ∣cos⁡3θ−8−4cos⁡2θ31111∣=0\left|\begin{array}{ccc}\cos 3 \theta & -8 & -4\\ \cos 2 \theta & 3 & 1\\ 1 & 1 & 1\end{array}\right|=0 (C1→C1−C2)&C2→C2−C3\left(\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}\right) \& \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3} ∣cos⁡3θ+8−4−3cos⁡2θ−321001∣=0\left|\begin{array}{ccc}\cos 3 \theta+8 & -4 & -3 \\ \cos 2 \theta-3 & 2 & 1\\ 0 & 0 & 1\end{array}\right|=0 2cos⁡3θ+16+4cos⁡2θ−12=02 \cos 3 \theta+16+4 \cos 2 \theta-12=0 (4cos⁡3θ−3cos⁡θ)+2(2cos⁡2θ−1)+2=0\left(4 \cos ^{3} \theta-3 \cos \theta\right)+2\left(2 \cos ^{2} \theta-1\right)+2=0 4cos⁡3θ+4cos⁡2θ−3cos⁡θ=04 \cos ^{3} \theta+4 \cos ^{2} \theta-3 \cos \theta=0 cos⁡θ(4cos⁡2θ+4cos⁡θ−3)=0\cos \theta\left(4 \cos ^{2} \theta+4 \cos \theta-3\right)=0 cos⁡θ(2cos⁡θ+3)(2cos⁡θ−1)=0\cos \theta(2 \cos \theta+3)(2 \cos \theta-1)=0 cos⁡θ=0,12,−32\cos \theta=0, \frac{1}{2},-\frac{3}{2} (rejected) θ=π2,3π2,π3,5π3\theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{\pi}{3}, \frac{5 \pi}{3} Sum⁡=4π\operatorname{Sum}=4 \pi

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Mathematics
Chapter
Determinants
Topic
System of Linear Equations using Determinants
The sum of all possible values of θ in [0, 2π ] , for which the… | JEE Main 2026 PYQ with Solution · DhiX AI