how can i find all the exact solutions to the equation : 2cos^2x + 3sinx = 3

the solutions have to be between [0,2pi)

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asked by Joe
  1. assuming that 2cos^2x means 2(cos(x))^2
    substitute using
    (cos(x))^2 = 1 - (sin(x))^2
    2(1 - (sin(x))^2) + 3sin(x) = 3
    2 - 2(sin(x))^2 + 3sin(x) = 3
    for clarity, let u = sin(x)
    2 - 2u^2 + 3u = 3
    2u^2 - 3 u + 1 = 0
    Solve this quadratic to find the values of sin x that are solutions. Finally, determine the angles in the range that have that value of sin(x).

    posted by Quidditch

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