The given equation 2cos2θ+3sinθ=0 can be written as
2(1−sin2θ)+3sinθ=0
⇒2sin2θ−3sinθ−2=0
⇒(2sinθ+1)(sinθ−2)=0
⇒sinθ=2,−21,(sinθ=2)
⇒sinθ=−21
Hence, solutions in [−2π,2π] are
−π+6π,−6π,π+6π,2π−6π
Sum of the elements is =2π.
JEE Main 2019 — Mathematics Trigonometry
Let S=θ∈[−2π,2π]:2cos2θ+3sinθ=0. Then the sum of the elements of S is:
Held on 9 Apr 2019 · Verified 6 Jul 2026.
π
613π
35π
2π
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