JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Trigonometric Equations

Group the terms of $2\sqrt{2}{\cos }^2\theta +(2-\sqrt{6})\cos \theta -\sqrt{3}=0$:



so $(\sqrt{2}\cos \theta +1)(2\cos \theta -\sqrt{3})=0$. Hence $\cos \theta =\frac{\sqrt{3}}{2}$ or $\cos \theta =-\frac{1}{\sqrt{2}}$. Each value lies strictly between $-1$ and $1$, so each yields two angles in every interval of length $2\pi$. The interval $[-2\pi ,2\pi ]$ spans two full periods, so each equation contributes four angles, and together they give $8$ solutions.