Number of solutions of √(3) cos 2 θ+8 cos θ+3 √(3)=0, θ in[-3 π, 2 π] is:

Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 23 January, Morning Shift — Question 16

Number of solutions of $\sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi]$ is:

$\sqrt{3}\left(2 \cos ^{2} \theta-1\right)+8 \cos \theta+3 \sqrt{3}=0$

$2 \sqrt{3} \cos ^{2} \theta+8 \cos \theta+2 \sqrt{3}=0$

$(\sqrt{3} \cos +1)(\cos \theta+\sqrt{3})=0$

$\cos \theta=-\frac{1}{\sqrt{3}}$ as $-\sqrt{3}$ (reject)

$\therefore \theta=$ will have 5 value in $[-3 \pi, 2 \pi]$

Ans. $=5 \Rightarrow$ option (2)

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Periodicity of trigometric functions,Solutions of trigonometric equations