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Math Help - trig functions

  1. #1
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    trig functions

    solve


    sqrt3cosxtanx+cosx where 0 less than or equal to x less than 2pi
    and

    cos2x-3sinx=2 where -pi is less than or equal to x less than or equal to pi

    exactly.




    what I tried was

    sqrt3cosx*sinx/cosx+cosx=0
    sqrt3sinx+cosx=0
    then I made each of them equal to 0
    sinx=
    0/sqrt3
    and
    cosx=0
    But that gives me the wrong answer



    And for the other one I tried a similar thing but I couldnt figure it out....
    thanks/
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by brentwoodbc View Post
    solve


    sqrt3cosxtanx+cosx where 0 less than or equal to x less than 2pi
    and

    cos2x-3sinx=2 where -pi is less than or equal to x less than or equal to pi

    exactly.




    what I tried was

    sqrt3cosx*sinx/cosx+cosx=0
    sqrt3sinx+cosx=0
    then I made each of them equal to 0
    sinx=
    0/sqrt3
    and
    cosx=0
    But that gives me the wrong answer


    And for the other one I tried a similar thing but I couldnt figure it out....
    thanks/
    I can't interpret the first one because you have grouped it funny, and it doesn't seem to be an equation at all!

    the second one however..........


    Which one should we use? well we probably would like everything to be in terms of sin so........

    1-2sin^2x+3sinx=2

    2sin^2x-3sinx+1=0

    you know how to solve quadratics right? (2x-1)(x-1)?......
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  3. #3
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    ya thanks that sort of works I get pi/6, 5pi/6 and pi/2 but the answer is pi/6,5pi/6 and pi/2 but all negatives :/

    if the restriction is -pi < x < pi with sine I thought it would be +



    also the other one was √3cosx*tanx+cosx=0
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  4. #4
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    Quote Originally Posted by brentwoodbc View Post
    also the other one was √3cosx*tanx+cosx=0
    You mean this?
    \sqrt{3}\cos x \tan x + cos x = 0

    If so, then...

    \begin{aligned}<br />
\sqrt{3}\cos x \tan x + \cos x &= 0 \\<br />
\cos x(\sqrt{3}\tan x + 1) &= 0<br />
\end{aligned}

    Set each factor equal to zero:

    \begin{aligned}<br />
\cos x &= 0 \\<br />
x &= \frac{\pi}{2}\;\;or\;\;\frac{3\pi}{2}<br />
\end{aligned}
    or
    \begin{aligned}<br />
\sqrt{3}\tan x + 1 &= 0 \\<br />
\sqrt{3}\tan x &= -1 \\<br />
\tan x &= -\frac{1}{\sqrt{3}}\;\;or\;\;-\frac{\sqrt{3}}{3} \\<br />
x &= \frac{5\pi}{6}\;\;or\;\;\frac{11\pi}{6}<br />
\end{aligned}


    01
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  5. #5
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    thanks that makes sense, and the way you wrote it was right,
    but I have a few problems

    1 Is why dont you factor out the cos in the tan "sinx/cosx"
    2 (more important) the answer given is 5pi/6 and 11pi/6 and if you graph it you only get the 2 answer's given?
    3 for the other problem I had do you know why the answers given are negative? I thought the pi/2 was + and the other two were negative?
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  6. #6
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    Quote Originally Posted by VonNemo19 View Post
    I can't interpret the first one because you have grouped it funny, and it doesn't seem to be an equation at all!

    the second one however..........


    Which one should we use? well we probably would like everything to be in terms of sin so........

    1-2sin^2x+3sinx=2

    2sin^2x-3sinx+1=0

    you know how to solve quadratics right? (2x-1)(x-1)?......
    how did

    1-2sin^2x+3sinx=2
    become a + ?
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  7. #7
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    Quote Originally Posted by brentwoodbc View Post
    1 Is why dont you factor out the cos in the tan "sinx/cosx"
    I don't understand what you're asking.
    2 (more important) the answer given is 5pi/6 and 11pi/6 and if you graph it you only get the 2 answer's given?
    Because I forgot to mention that x = \frac{\pi}{2}\;\;or\;\;\frac{3\pi}{2} are extraneous solutions -- if you plug either into the original equation tan x is undefined.


    01
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  8. #8
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    you jerk


    seriously thanks.
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