• May 11th 2011, 02:16 AM
domenfrandolic
Hello!

I have to solve this

tan 3θ = 0

what is then θ

there should be three values...

but my range is between -pi to +pi
• May 11th 2011, 02:31 AM
Prove It
What have you tried?
• May 11th 2011, 04:08 AM
domenfrandolic
my teacher said i came to that point...so it is OK...but from noe on i am stuck
• May 11th 2011, 04:53 AM
Soroban
Hello, domenfrandolic!

There are five solutions.

Quote:

$\displaystyle \text{Solve: }\;\tan 3\theta \:=\: 0\quad\text{ for }\text{-}\pi < \theta < +\pi$

We have: .$\displaystyle \tan3\theta \:=\:0$

Then: .$\displaystyle 3\theta \:=\:\hdots\;\text{-}4\pi,\;\text{-}3\pi,\;\text{-}2\pi,\;\text{-}\pi,\;0,\;\pi,\;2\pi,\;3\pi,\;4\pi\;\hdots$

Hence: .$\displaystyle \theta \;=\;\hdots\;\text{-}\tfrac{4\pi}{3},\;\text{-}\pi,\;\text{-}\tfrac{2\pi}{3},\;\text{-}\tfrac{\pi}{3},\;0,\;\tfrac{\pi}{3},\;\tfrac{2\pi }{3},\;\pi,\;\tfrac{4\pi}{3}\;\hdots$

Therefore: .$\displaystyle \theta \;=\;\left\{\text{-}\frac{2\pi}{3},\;\text{-}\frac{\pi}{3},\;0,\;\frac{\pi}{3},\;\frac{2\pi}{3 }\right\}$