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Thread: 3tanx + 2cosx = 0 Solving for x but confused

  1. #1
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    Exclamation 3tanx + 2cosx = 0 Solving for x but confused

    Okay, so, I have been given this question to solve for x and get the possible graph solutions. But I looked at part of the answer and don't understand how some of it works.

    So the answer is :

    3(sinx / cosx) + 2cosx = 0
    3sinx + 2cosx / cosx = 0

    But then the next line just goes to

    3sinx + 2cosx = 0

    How can you divide it by cosx and not change the top? I'm so confused!
    And then I don't know where I would go from here?
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  2. #2
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    Re: 3tanx + 2cosx = 0 Solving for x but confused

    Consider the following analogy
    x /2 = 0
    x = 0

    Note that the domain of tan(x) doesn't include point where cos(x)=0, so you're multiplying both sides by a non-zero number when you multiply both sides by cos(x)

    Misread. Should be this before you conclude the numerator of LHS is 0.
    3sin(x) + 2cos^2(x) / cos(x) = 0
    Last edited by MacstersUndead; Sep 25th 2018 at 06:13 AM.
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  3. #3
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    Re: 3tanx + 2cosx = 0 Solving for x but confused

    Quote Originally Posted by smurfledurple View Post
    Okay, so, I have been given this question to solve for x and get the possible graph solutions. But I looked at part of the answer and don't understand how some of it works.

    So the answer is :

    3(sinx / cosx) + 2cosx = 0
    3sinx + 2cosx / cosx = 0

    But then the next line just goes to

    3sinx + 2cosx = 0
    What you wrote doesn't become that! But (3sinx+ 2cosx)/cosx= 0 does! Those parentheses are important!

    How can you divide it by cosx and not change the top? I'm so confused!
    And then I don't know where I would go from here?
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  4. #4
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    Re: 3tanx + 2cosx = 0 Solving for x but confused

    $3\sin x+2\cos^2 x=0$

    $3\sin x + 2(1-\sin^2 x)=0$

    Let $u=\sin x$

    $2u^2-3u-2=0$

    $(2u+1)(u-2)=0$

    Since there are no solutions to $\sin x=2$ it must be that $\sin x =-\dfrac 1 2$.

    So $x=\dfrac{7\pi}{6}+2n\pi$ or $x=\dfrac{11\pi}{6}+2n\pi$.
    Thanks from HallsofIvy
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