JEE Main 2025 Apr 2 Shift 2, Mathematics Q9: Trigonometric Equations

Treat $\mathrm{cosec}\theta$ as the unknown. Using the quadratic formula on $\sqrt{3}{t}^2-2(\sqrt{3}-1)t-4=0$:

$t=\frac{2(\sqrt{3}-1)\pm \sqrt{4(\sqrt{3}-1{)}^2+16\sqrt{3}}}{2\sqrt{3}}=\frac{2(\sqrt{3}-1)\pm \sqrt{16+8\sqrt{3}}}{2\sqrt{3}}$.

Since $16+8\sqrt{3}=(2+2\sqrt{3}{)}^2$, the square root is $2+2\sqrt{3}$. Hence:

$t=\frac{2(\sqrt{3}-1)\pm (2+2\sqrt{3})}{2\sqrt{3}}\Rightarrow t=2\text{or}t=-\frac{2}{\sqrt{3}}$.

So $\mathrm{cosec}\theta =2\Rightarrow \sin \theta =\frac{1}{2}$, or $\mathrm{cosec}\theta =-\frac{2}{\sqrt{3}}\Rightarrow \sin \theta =-\frac{\sqrt{3}}{2}$.

The interval $\left[-\frac{7\pi }{6},\frac{4\pi }{3}\right]$ has length $\frac{5\pi }{2}$, slightly more than one full period. Each equation $\sin \theta =c$ (with $|c|<1$) contributes $3$ solutions in this interval, giving $3+3=6$ solutions in total.