# Thread: Solving Trigonometric equations 2

1. ## Solving Trigonometric equations 2

Question
$\displaystyle tan^3x-tanx=0$ if $\displaystyle 0 \leq x \leq \frac{\pi}{2}$

Attempt

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

Thank you

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

Attempt

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

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 $\displaystyle \left[0,\frac{\pi}{2}\right]$

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

Attempt

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

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 $\displaystyle \frac{\pi}{2}$