JEE Main 2026 23 January Morning Shift
Exam JEE Main
Year 2026
Date 23 January
Shift Morning Shift
Subject Mathematics
Chapter Trigonometric Equations JEE Main 2026 · 23 January · Morning Shift · Mathematics · Trigonometric Equations
Number of solutions of 3 2 θ+8 θ+3 3 =0, θ ∈[-3 π, 2 π] is:
Number of solutions of
3 cos 2 θ + 8 cos θ + 3 3 = 0 , θ ∈ [ − 3 π , 2 π ] \sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi] 3 cos 2 θ + 8 cos θ + 3 3 = 0 , θ ∈ [ − 3 π , 2 π ] is:
Step-by-step solution
Using
cos 2 θ = 2 cos 2 θ − 1 \cos 2\theta = 2\cos^2\theta - 1 cos 2 θ = 2 cos 2 θ − 1 :
2 3 cos 2 θ + 8 cos θ + 2 3 = 0 ⇒ 3 cos 2 θ + 4 cos θ + 3 = 0 2\sqrt{3}\cos^2\theta + 8\cos\theta + 2\sqrt{3} = 0 \Rightarrow \sqrt{3}\cos^2\theta + 4\cos\theta + \sqrt{3} = 0 2 3 cos 2 θ + 8 cos θ + 2 3 = 0 ⇒ 3 cos 2 θ + 4 cos θ + 3 = 0 .
cos θ = − 4 ± 2 2 3 \cos\theta = \frac{-4 \pm 2}{2\sqrt{3}} cos θ = 2 3 − 4 ± 2 , giving
cos θ = − 1 3 \cos\theta = -\frac{1}{\sqrt{3}} cos θ = − 3 1 or
cos θ = − 3 \cos\theta = -\sqrt{3} cos θ = − 3 (rejected).
Let
α = cos − 1 ( − 1 / 3 ) ≈ 2.186 \alpha = \cos^{-1}(-1/\sqrt{3}) \approx 2.186 α = cos − 1 ( − 1/ 3 ) ≈ 2.186 rad. General solution:
θ = 2 n π ± α \theta = 2n\pi \pm \alpha θ = 2 nπ ± α .
In
[ − 3 π , 2 π ] [-3\pi, 2\pi] [ − 3 π , 2 π ] :
θ = α , − α , 2 π − α , − 2 π + α , − 2 π − α \theta = \alpha, -\alpha, 2\pi-\alpha, -2\pi+\alpha, -2\pi-\alpha θ = α , − α , 2 π − α , − 2 π + α , − 2 π − α .
Total
= 5 = 5 = 5 solutions.
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