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5 January, 07:09

Solve on the interval [0,2pi) 2sin^2x-3sinx+1=0

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  1. 5 January, 07:22
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    2sin² (x) + 3sin (x) + 1 = 0

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

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

    sin (x) = - 1

    x = {3π/2}

    2sin (x) + 1 = 0

    sin (x) = - 1/2

    x = {7π/6, 11π/6}

    x = {7π/6, 3π/2, 11π/6}
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