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Math Help - Trig Equations

  1. #1
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    Trig Equations

    Solve the following equations for 0<x<2pi

    4cosx(cosx-1)=-5cosx

    i managed to do-

    4cos^2(x) - 4cosx = -5cosx
    4cos^2(x) + cosx = 0
    4cosx + 1 = 0
    cosx = -1/4

    x = 104.5

    I used the cos graph but the answers I was given were-

    x= 90, -90, 104, -104

    bit confused
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    Quote Originally Posted by greghunter View Post
    i managed to do-

    4cos^2(x) - 4cosx = -5cosx
    4cos^2(x) + cosx = 0
    4cosx + 1 = 0
    cosx = -1/4

    x = 104.5

    I used the cos graph but the answers I was given were-

    x= 90, -90, 104, -104
    You have one of the answers, but you lost the other when you divided by \cos x in step 3.

    From 4\cos^2x + \cos x = 0 we can factor

    \cos x(4\cos x+1)=0\text.

    Equating the two factors with zero gives two separate equations:

    \cos x=0 (solution on (0,2\pi) is x=\pi/2)

    and

    4\cos x+1=0

    \Rightarrow\cos x=-\frac14 (solution on (0,2\pi) is x=\arccos(-1/4)\approx1.8235\approx104.48^\circ).

    There should not be any negative solutions, since it was stated that 0<x<2\pi\text.
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