JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Determinants

A non-trivial solution requires the coefficient determinant to vanish: $\left|\begin{array}{ccc}\cos 3\theta & -8& -12\\ \cos 2\theta & 3& 3\\ 1& 1& 3\end{array}\right|=0$, which expands to $6\cos 3\theta +12\cos 2\theta +12=0$, i.e. $\cos 3\theta +2\cos 2\theta +2=0$. Using $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$ and $\cos 2\theta =2{\cos }^2\theta -1$, this becomes $4{\cos }^3\theta +4{\cos }^2\theta -3\cos \theta =0$, i.e. $\cos \theta (2\cos \theta -1)(2\cos \theta +3)=0$. Rejecting $\cos \theta =-\frac{3}{2}$, the solutions in $[0,2\pi ]$ are $\theta =\frac{\pi }{2},\frac{3\pi }{2}$ from $\cos \theta =0$ and $\theta =\frac{\pi }{3},\frac{5\pi }{3}$ from $\cos \theta =\frac{1}{2}$. Their sum is $\frac{\pi }{2}+\frac{3\pi }{2}+\frac{\pi }{3}+\frac{5\pi }{3}=4\pi$.