JEE Main 2026 — Mathematics Algebra

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

$$\begin{vmatrix} \cos 3\theta & -8 & -12 \\ \cos 2\theta & 3 & 3 \\ 1 & 1 & 3 \end{vmatrix} = 0$$​cos3θcos2θ1​−831​−1233​​=0

Expanding the determinant along the first row:

$$\cos 3\theta (9 - 3) - (-8)(3\cos 2\theta - 3) + (-12)(\cos 2\theta - 3) = 0$$cos3θ(9−3)−(−8)(3cos2θ−3)+(−12)(cos2θ−3)=0

$$6\cos 3\theta + 24\cos 2\theta - 24 - 12\cos 2\theta + 36 = 0$$6cos3θ+24cos2θ−24−12cos2θ+36=0

$$6\cos 3\theta + 12\cos 2\theta + 12 = 0$$6cos3θ+12cos2θ+12=0

Dividing by $6$6, we get:

$$\cos 3\theta + 2\cos 2\theta + 2 = 0$$cos3θ+2cos2θ+2=0

Using the multiple angle formulas $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$cos3θ=4cos3θ−3cosθ and $\cos 2\theta = 2\cos^2 \theta - 1$cos2θ=2cos2θ−1, we substitute these into the equation:

$$(4\cos^3 \theta - 3\cos \theta) + 2(2\cos^2 \theta - 1) + 2 = 0$$(4cos3θ−3cosθ)+2(2cos2θ−1)+2=0

$$4\cos^3 \theta + 4\cos^2 \theta - 3\cos \theta = 0$$4cos3θ+4cos2θ−3cosθ=0

$$\cos \theta (4\cos^2 \theta + 4\cos \theta - 3) = 0$$cosθ(4cos2θ+4cosθ−3)=0

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

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

$$\cos \theta = 0 \implies \theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$$cosθ=0⟹θ=2π​,23π​

$$\cos \theta = \dfrac{1}{2} \implies \theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}$$cosθ=21​⟹θ=3π​,35π​

$\cos \theta = -\dfrac{3}{2}$cosθ=−23​ (Not possible since $\cos \theta \in [-1, 1]$cosθ∈[−1,1])

The possible values of $\theta \in [0, 2\pi]$θ∈[0,2π] are $\dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{\pi}{3}, \dfrac{5\pi}{3}$2π​,23π​,3π​,35π​.

Sum of all possible values of $\theta$θ:

$$\sum \theta = \left(\dfrac{\pi}{2} + \dfrac{3\pi}{2}\right) + \left(\dfrac{\pi}{3} + \dfrac{5\pi}{3}\right) = 2\pi + 2\pi = 4\pi$$∑θ=(2π​+23π​)+(3π​+35π​)=2π+2π=4π

Answer: $4\pi$4π