Solving Trigonometric equations 2

• Mar 23rd 2009, 11:03 AM
mj.alawami
Solving Trigonometric equations 2
Question
$\displaystyle tan^3x-tanx=0$ if $\displaystyle 0 \leq x \leq \frac{\pi}{2}$

Attempt

$\displaystyle tan(tan+1)(t-1)=0$(Crying)

Thank you
• Mar 23rd 2009, 11:19 AM
running-gag
Quote:

Originally Posted by mj.alawami
Question
$\displaystyle tan^3x-tanx=0$ if $\displaystyle 0 \leq x \leq \frac{\pi}{2}$

Attempt

$\displaystyle tan(tan+1)(t-1)=0$(Crying)

Thank you

Hi

That's OK
$\displaystyle \tan x (\tan x+1)(\tan x-1)=0$

$\displaystyle \tan x = 0 \rightarrow x=0$
$\displaystyle \tan x = 1 \rightarrow x=\frac{\pi}{4}$
$\displaystyle \tan x = -1 \rightarrow$ no solution in $\displaystyle \left[0,\frac{\pi}{2}\right]$
• Mar 23rd 2009, 11:20 AM
e^(i*pi)
Quote:

Originally Posted by mj.alawami
Question
$\displaystyle tan^3x-tanx=0$ if $\displaystyle 0 \leq x \leq \frac{\pi}{2}$

Attempt

$\displaystyle tan(tan+1)(t-1)=0$(Crying)

Thank you

1.Factor out tan(x) to get $\displaystyle tan(x)=0$

Now solve the quadratic for the other solutions remembering there is an asymptote at $\displaystyle \frac{\pi}{2}$