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Mathematics · Trigonometry Ratios and Identities

JEE Main 2025 — 2 April, Evening Shift — Question 26

If θ∈[−7π6,4π3]\theta \in\left[-\frac{7 \pi}{6}, \frac{4 \pi}{3}\right], then the number of solutions of 3cosec⁡2θ−2(3−1)cosec⁡θ−4=0\sqrt{3} \operatorname{cosec}^{2} \theta-2(\sqrt{3}-1) \operatorname{cosec} \theta-4=0, is

equal to

  1. Option A:

    8

  2. Option B:

    10

  3. Option C:

    7

  4. Option D:

    6

    Correct

Answer: D

Step-by-step solution

cosec⁡θ=23−2±(2(3−1))2+43(4)2323−2±4(3+1−23)+16323=23−2±2(3+1)23⇒cosec⁡θ=−23,2θ=−7π6,−2π3,−π3,π6,5π6,4π3\begin{aligned} & \operatorname{cosec} \theta=\frac{2 \sqrt{3}-2 \pm \sqrt{(2(\sqrt{3}-1))^{2}+4 \sqrt{3}(4)}}{2 \sqrt{3}} \\& \frac{2 \sqrt{3}-2 \pm \sqrt{4(3+1-2 \sqrt{3})+16 \sqrt{3}}}{2 \sqrt{3}} \\& \frac{=2 \sqrt{3}-2 \pm 2(\sqrt{3}+1)}{2 \sqrt{3}} \\& \Rightarrow \operatorname{cosec} \theta=\frac{-2}{\sqrt{3}}, 2 \\& \theta=\frac{-7 \pi}{6}, \frac{-2 \pi}{3}, \frac{-\pi}{3}, \frac{\pi}{6}, \frac{5 \pi}{6}, \frac{4 \pi}{3} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
2 April, Evening Shift
Subject
Mathematics
Chapter
Trigonometry Ratios and Identities
Topic
Periodicity of trigometric functions,Solutions of trigonometric equations