math

solve 2cos^2x-5cosx+2=0 for principal values of x
I don't understand this question...what are principal values?

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  1. those solutions where 0 ≤ x ≤ 2π

    2cos^2x-5cosx+2=0
    this is a standard quadratic where your variable is cosx
    How about doing this ?
    let t = cosx
    then we have
    2t^2 - 5t + 2 = 0 , which factors to
    (2t - 1)(t - 2) = 0
    t = 1/2 or t = 2

    so cosx = 1/2 or cosx = 2
    that last part is silly, since cos(anything) lies between -1 and +1

    so cosx = +1/2, which tells me that the angle x must be in quads I or IV by the CAST rule
    we know that x = 60° or π/3
    We are in I or IV
    so x = 60 = 60° or π/3
    or x = 360-60 = 300° or 5π/3

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