JEE Main 2026 — Determinants Question with Solution

For the given system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

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

Expanding the determinant along the first row:

$\cos 3\theta (9-3)-(-8)(3\cos 2\theta -3)+(-12)(\cos 2\theta -3)=0$

$6\cos 3\theta +24\cos 2\theta -24-12\cos 2\theta +36=0$

$6\cos 3\theta +12\cos 2\theta +12=0$

Dividing by $6$, we get:

$\cos 3\theta +2\cos 2\theta +2=0$

Using the multiple angle formulas $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$ and $\cos 2\theta =2{\cos }^2\theta -1$, we substitute these into the equation:

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

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

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

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

This gives the possible values for $\cos \theta$:

$\cos \theta =0⟹\theta =\frac{\pi }{2},\frac{3\pi }{2}$

$\cos \theta =\frac{1}{2}⟹\theta =\frac{\pi }{3},\frac{5\pi }{3}$

$\cos \theta =-\frac{3}{2}$ (Not possible since $\cos \theta \in [-1,1]$)

The possible values of $\theta \in [0,2\pi ]$ are $\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{5\pi }{3}$.

Sum of all possible values of $\theta$:

$\sum \theta =\left(\frac{\pi }{2}+\frac{3\pi }{2}\right)+\left(\frac{\pi }{3}+\frac{5\pi }{3}\right)=2\pi +2\pi =4\pi$

Answer: $4\pi$