Q49 / 75 · JEE Main 2025 (2 Apr Shift 2) · Mathematics
If $\theta \in\left[-\frac{7 \pi}{6}, \frac{4 \pi}{3}\right]$, then the number of solutions of $\sqrt{3} \operatorname{cosec}^2 \theta-2(\sqrt{3}-1) \operatorname{cosec} \theta-4=0$, is equal to
- Correct answerA$6$
- B$8$
- C$10$
- D$7$
Solution
$\begin{aligned} & \operatorname{cosec} \theta=\frac{2(\sqrt{3}-1) \pm \sqrt{4(3+1-2 \sqrt{3})+16 \sqrt{3}}}{2 \sqrt{3}} \\ & =\frac{2(\sqrt{3-1}) \pm \sqrt{16+8 \sqrt{3}}}{2 \sqrt{3}} \\ & =\frac{2(\sqrt{3}-1) \pm(2+2 \sqrt{3})}{2 \sqrt{3}}\end{aligned}$
$\operatorname{cosec} \theta=2$ or $\frac{-2}{\sqrt{3}}$
$\therefore \sin \theta=\frac{1}{2}$ or $\frac{-\sqrt{3}}{2}$
$\therefore \sin \theta=\frac{1}{2}$ has 3 solutions \& also $\sin \theta=\frac{-\sqrt{3}}{2}$
has 3 solutions in $\left[\frac{-7 \pi}{6}, \frac{4 \pi}{3}\right]$
$\operatorname{cosec} \theta=2$ or $\frac{-2}{\sqrt{3}}$
$\therefore \sin \theta=\frac{1}{2}$ or $\frac{-\sqrt{3}}{2}$
$\therefore \sin \theta=\frac{1}{2}$ has 3 solutions \& also $\sin \theta=\frac{-\sqrt{3}}{2}$
has 3 solutions in $\left[\frac{-7 \pi}{6}, \frac{4 \pi}{3}\right]$
Hence, the correct answer is option: 'A'.