JEE Main 2026 Apr 6 Shift 2, Mathematics Q4: Linear Equations And Cramer’s Rule

Non-trivial solutions exist iff the coefficient determinant is zero:

$\left|\begin{array}{c}\cos 3\theta -8-12\\ \cos 2\theta 33\\ 113\end{array}\right|=0$.

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

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

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

Substitute $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$ and $\cos 2\theta =2{\cos }^2\theta -1$. Dividing by 2:

$3(4{\cos }^3\theta -3\cos \theta )+6(2{\cos }^2\theta -1)+6=0$

$\Rightarrow 12{\cos }^3\theta +12{\cos }^2\theta -9\cos \theta =0$ $\Rightarrow 3\cos \theta (4{\cos }^2\theta +4\cos \theta -3)=0$.

$\cos \theta =0$ or $(2\cos \theta -1)(2\cos \theta +3)=0$ $\Rightarrow \cos \theta =0,\frac{1}{2}$ ($-\frac{3}{2}$ rejected).

In $[0,2\pi ]$: $\theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{5\pi }{3}$.

Sum $=\frac{\pi }{2}+\frac{3\pi }{2}+\frac{\pi }{3}+\frac{5\pi }{3}=2\pi +2\pi =4\pi$.