JEE Main 2025 Jan 22 Shift 2, Mathematics Q18: Trigonometric Equations

Equation 1: $2{\sin }^2\theta =\cos 2\theta =1-2{\sin }^2\theta$, so $4{\sin }^2\theta =1$, giving $\sin \theta =\pm \frac{1}{2}$.

Equation 2: $2{\cos }^2\theta =3\sin \theta$, i.e. $2(1-{\sin }^2\theta )=3\sin \theta$, so $2{\sin }^2\theta +3\sin \theta -2=0$. Factor: $(2\sin \theta -1)(\sin \theta +2)=0$. Since $\sin \theta \ne -2$, we get $\sin \theta =\frac{1}{2}$.

Common solution: $\sin \theta =\frac{1}{2}$. In $[0,2\pi ]$ this gives $\theta =\frac{\pi }{6}$ and $\theta =\frac{5\pi }{6}$.

Sum $=\frac{\pi }{6}+\frac{5\pi }{6}=\pi$.