Mathematics · 2025
JEE Main · 2 April 2025, Shift 2 · Q13
If θ ∈[-(7 π)/6, (4 π)/3], then the number of solutions of √ 3 cosec^2 θ-2(√ 3 -1) cosec θ-4=0, is equal to:
If $\displaystyle \theta \in\left[-\frac{7 \pi}{6}, \frac{4 \pi}{3}\right]$, then the number of solutions of $\displaystyle \sqrt{3} \operatorname{cosec}^2 \theta-2(\sqrt{3}-1) \operatorname{cosec} \theta-4=0$, is equal to :
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 6$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.