JEE Main 2025 Apr 2 Shift 1, Mathematics Q7: Trigonometric Equations

Split the linear coefficient $(2-\sqrt{6})=2-\sqrt{6}$ to factor:

$$2\sqrt{2}{\cos }^2\theta +2\cos \theta -\sqrt{6}\cos \theta -\sqrt{3}=0$$

Group the terms:

$$2\cos \theta (\sqrt{2}\cos \theta +1)-\sqrt{3}(\sqrt{2}\cos \theta +1)=0$$

$$(\sqrt{2}\cos \theta +1)(2\cos \theta -\sqrt{3})=0$$

So:

$$\cos \theta =-\frac{1}{\sqrt{2}}\text{or}\cos \theta =\frac{\sqrt{3}}{2}$$

Counting solutions in $[-2\pi ,2\pi ]$: over an interval of length $4\pi$, each non-extreme value of $\cos \theta$ is attained 4 times.

$\cos \theta =-\frac{1}{\sqrt{2}}$: $4$ solutions.

$\cos \theta =\frac{\sqrt{3}}{2}$: $4$ solutions.

Total $=4+4=8$.

$\therefore$ Option (3) is correct.