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13 June, 17:33

What are all the exact solutions of 2sin^2x-sinx=0 for 0

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  1. 13 June, 18:12
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    x = {0, π/6}

    Step-by-step explanation:

    2sin^2x-sinx=0

    and i know the answer is this ...

    2sin² (x) - sin (x) = 0

    sin2 (x) is sin (x) * sin (x) so 2sin² (x) - sin (x) = 2 sin (x) * sin (x) - sin (x) and this expression has a common factor of sin (x) so

    2sin² (x) - sin (x) = 2 sin (x) * sin (x) - sin (x) = sin (x) [2 sin (x) - 1]

    sin (x) (2sin (x) - 1) = 0

    sin (x) = 0

    x = 0

    2sin (x) - 1 = 0

    sin (x) = 1/2

    x = π/6

    x = {0, π/6}
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