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Thread: trig problem

  1. #1
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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
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    ok from your first step
    $\displaystyle

    2cosX + \frac{sinX}{cosX} = \frac{1}{cosX}
    $

    mult thru with $\displaystyle cosX$

    $\displaystyle 2cos^2X + sinX = 1$

    since

    $\displaystyle cos^2X = 1 - sin^2X$

    then

    $\displaystyle 2(1 - sin^2X) + sinX = 1$

    $\displaystyle 2sin^x - sinX - 1 = 0$

    $\displaystyle (2sinx + 1)(sinX - 1)$

    $\displaystyle sinX = -\frac{1}{2}$ and $\displaystyle sin^{-1}(-\frac{1}{2})$ so x = $\displaystyle -\frac{\pi}{6}$ or $\displaystyle \frac{11\pi}{6}$
    Last edited by bigwave; Dec 3rd 2009 at 02:22 PM. Reason: more info
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  3. #3
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    then what?
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