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?
$\displaystyle f(x) = g(x)$
$\displaystyle 1 - \tan^{2}x = \tan x$
$\displaystyle 0 = tan^{2}x + tanx - 1$
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.
$\displaystyle 1-\tan^2(x)=\tan(x) \iff \tan^2(x)+tan(x)-1=0$
$\displaystyle u=tan(x)$ 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.