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Thread: Trignometric Equations

  1. #1
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    Trignometric Equations

    Let f(x)=1-tan^2x and g(x)=tanx. Find the value of x in the interval (0, pie/2) where f(x)=g(x)

    Can anyone help me with this question?
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  2. #2
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    $\displaystyle f(x) = g(x)$
    $\displaystyle 1 - \tan^{2}x = \tan x$
    $\displaystyle 0 = tan^{2}x + tanx - 1$

    You can see that you can solve it as if it were a quadratic equation. Just make sure your solution for x is in the solution set provided.

    Edit: Just to make it easier to visualize, let y = tan x. So you have $\displaystyle 0 = y^{2} + y - 1$. So:

    $\displaystyle y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

    $\displaystyle y = \frac{-1 \pm \sqrt{(-1)^{2} - 4(1)(-1)}}{2(1)}$

    etc.
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by lemontea View Post
    Let f(x)=1-tan^2x and g(x)=tanx. Find the value of x in the interval (0, pie/2) where f(x)=g(x)

    Can anyone help me with this question?

    $\displaystyle 1-\tan^2(x)=\tan(x) \iff \tan^2(x)+tan(x)-1=0$

    $\displaystyle u=tan(x)$ we get..

    $\displaystyle u^2+u-1=0$ by the quadratic formula we get..

    $\displaystyle tan(x) = u= \frac{-1 \pm \sqrt{1-4(1)(-1)}}{2(1)}=\frac{-1 \pm \sqrt{5}}{2}$

    You should be able to finish from here. Good luck.
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  4. #4
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    the answer i got is

    y=-1+squareroot 5/2 or y=-1-squareroot 5/2

    am i right?
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  5. #5
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    Yes. So:

    $\displaystyle \tan x = y = \frac{-1 \pm \sqrt{5}}{2}$

    $\displaystyle x = \tan^{-1} \left(\frac{-1 + \sqrt{5}}{2}\right) \quad \quad x = \tan^{-1} \left(\frac{-1 - \sqrt{5}}{2}\right)$
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