JEE 2025 — Mathematics PYQ

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

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

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

In $[0,2\pi ]$, $\cos \theta =\frac{\sqrt{3}}{2}$, then $\theta =\pm \frac{\pi }{6}+\frac{11\pi }{6}$

If $\cos \theta =-\frac{1}{\sqrt{2}}$, then $\theta =\pm \frac{3\pi }{4},\pm \frac{5\pi }{4}$

Hence, there are a total of 8 solutions because cosx is an even function.