JEE Main 2020 — Mathematics Trigonometry

From $\cos A + \cos B = 0$: $\cos B = -\cos A$, so $B = \pi - A$

$\sin A + \sin(\pi - A) = 2\sin A = 1 \Rightarrow \sin A = \frac{1}{2}$

$A = \frac{\pi}{6}$, $\cos 2A = \cos\frac{\pi}{3} = \frac{1}{2}$

$B = \frac{5\pi}{6}$, $\cos 2B = \cos\frac{5\pi}{3} = \frac{1}{2}$

$$12\cos 2A + 4\cos 2B = 12 \cdot \frac{1}{2} + 4 \cdot \frac{1}{2} = 6 + 2 = 8$$