JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 4

Non-trivial solution needs a zero determinant:
$$\begin{vmatrix}\cos3\theta&-8&-12\\\cos2\theta&3&3\\1&1&3\end{vmatrix}=6\cos3\theta+12\cos2\theta+12=0.$$

In $c=\cos\theta$. $4c^3-3c+2(2c^2-1)+2=c(4c^2+4c-3)=c(2c-1)(2c+3)=0$, so $c=0$ or $\tfrac12$.

Angles in $[0,2\pi]$. $\tfrac\pi2,\tfrac{3\pi}2,\tfrac\pi3,\tfrac{5\pi}3$; their sum is $4\pi$.

(D)