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Math Help - trig problem

  1. #1
    Newbie
    Joined
    Nov 2009
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    trig problem

    Here's the problem:

    2cos x + tan x= sec x

    I changed the tan x into sin x/cos x

    I changed the sec x into 1/cos x

    Then I multiplied by cos x and then I factored. I got (2sin x + 3)(3sin x-1)=0

    sin x = -3/2 and sin x = 1

    But that's not the answer on the answer sheet. What did I do wrong?
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  2. #2
    Super Member bigwave's Avatar
    Joined
    Nov 2009
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    honolulu
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    ok from your first step
     <br /> <br />
2cosX + \frac{sinX}{cosX} = \frac{1}{cosX}<br />

    mult thru with cosX

    2cos^2X + sinX = 1

    since

    cos^2X = 1 - sin^2X

    then

    2(1 - sin^2X) + sinX = 1

    2sin^x - sinX - 1 = 0

    (2sinx + 1)(sinX - 1)

    sinX = -\frac{1}{2} and sin^{-1}(-\frac{1}{2}) so x = -\frac{\pi}{6} or \frac{11\pi}{6}
    Last edited by bigwave; December 3rd 2009 at 03:22 PM. Reason: more info
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  3. #3
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    then what?
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