Let S=θ∈[−2π,2π]:2cos2⁡θ+3sin⁡θ=0Then the sum of the elements of S is :

Let S=θ[2π,2π]:2cos2θ+3sinθ=0Then the sum of the elements of S is :

  1. A

    2π

  2. B

    π

  3. C

    5π/3

  4. D

    13π/6

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    Solution:

    Given, 2cos2θ+3sinθ=0

    21sin2θ+3sinθ=022sin2θ+3sinθ=02sin2θ4sinθ+sinθ2=02sinθ(sinθ2)+1(sinθ2)=0(2sinθ+1)(sinθ2)=0

    sinθ=12                   [ sinθ20]

     θ=π6,5π6,7π6,11π6

    Required sum =π5π+7π+11π6=12π6=2π

     

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