JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Trigonometric Equations

Writing $\tan 2\theta =\frac{\sin 2\theta }{\cos 2\theta }$,

The factor $2{\cos }^2\theta$ is positive except at $\theta =\frac{\pi }{2}$ and $\theta =\frac{3\pi }{2}$, where the whole expression vanishes, so the inequality holds precisely where $\tan 2\theta >0$. For $\theta \in [0,2\pi ]$ the angle $2\theta$ runs over $[0,4\pi ]$, and $\tan 2\theta >0$ on $\left(0,\frac{\pi }{2}\right)\cup \left(\pi ,\frac{3\pi }{2}\right)\cup \left(2\pi ,\frac{5\pi }{2}\right)\cup \left(3\pi ,\frac{7\pi }{2}\right)$. Halving these intervals, the values of $\theta$ are $\left(0,\frac{\pi }{4}\right)\cup \left(\frac{\pi }{2},\frac{3\pi }{4}\right)\cup \left(\pi ,\frac{5\pi }{4}\right)\cup \left(\frac{3\pi }{2},\frac{7\pi }{4}\right)$.