Using cos2θ=2cos2θ−1:
23cos2θ+8cosθ+23=0⇒3cos2θ+4cosθ+3=0.
cosθ=23−4±2, giving cosθ=−31 or cosθ=−3 (rejected).
Let α=cos−1(−1/3)≈2.186 rad. General solution: θ=2nπ±α.
In [−3π,2π]: θ=α,−α,2π−α,−2π+α,−2π−α.
Total =5 solutions.
JEE Main 2026 — Mathematics Trigonometry
Number of solutions of 3cos2θ+8cosθ+33=0,θ∈[−3π,2π] is:
Held on 23 Jan 2026 · Verified 6 Jul 2026.
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5
3
4
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