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4 June, 21:20

What is the solution to the trigonometric inequality 2sin (x) + 3> sin^2 (x)

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Answers (1)
  1. 4 June, 21:44
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    First of all, let's analyze this equation. There is no restriction for

    x since the domain of a function sin

    (x)

    is all real numbers.

    Secondly, we can introduce an intermediary variable y = sin (x

    and express this equation in terms of y : y 2 2 y + 3 or y 2 - 2 y 3 = 0

    This equation has solutions: y 1, 2 = 2 ± √ 4 + 12 2 or y 1 = 3 and 2 = - 1

    Now we have to take into consideration that y = sin (x)

    and, as such, is restricted in values from - 1 to 1

    (inclusive).

    Therefore, solution y 1 = 3

    would not lead to any value of x.

    The second solution y 2 - 1

    can be used to find x : sin (x) = 1

    implies that x = - π 2 + 2 π N, where N - any integer number or, in degrees, x = -

    90 o +

    360 o ⋅

    N

    This is a final solution to the problem
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