Let θ∈0,4π satisfy the equation sinθ+2sinθ+3sinθ+4=6 .If the sum of all the values of θ is of the form kπ , then the value of k is

Let θ0,4π satisfy the equation sinθ+2sinθ+3sinθ+4=6 .If the sum of all the values of θ is of the form kπ , then the value of k is

  1. A

    6

  2. B

    5

  3. C

    4

  4. D

    2

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    Solution:

    We have sinθ+2sinθ+3sinθ+4=6

    Since -1sinθ1, LHS6

    But RHS=6sinθ=-1

    Since θ0,4πθ=3π2,7π2

    sum of values of θ=3π2+7π2=5π= k=5

     

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