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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- Mar 18th 2008, 07:41 PMlemonteaTrignometric 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? - Mar 18th 2008, 07:46 PMo_O
$\displaystyle f(x) = g(x)$

$\displaystyle 1 - \tan^{2}x = \tan x$

$\displaystyle 0 = tan^{2}x + tanx - 1$

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. - Mar 18th 2008, 07:52 PMTheEmptySet

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

$\displaystyle u=tan(x)$ 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. - Mar 18th 2008, 08:01 PMlemontea
the answer i got is

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

am i right? - Mar 18th 2008, 08:10 PMo_O
Yes. So:

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

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