Question

Help me solve this? I'll give a medal and fan, ha.

tan(3x)(tan x -1) = 0
interval [0,2p)

Answer 1

Since you have 2 factors, then you have that
\(\tan(3x)=0\) or \(\tan x-1=0 \implies \tan x=1\)
This implies:
\(\arctan(0)=3x\) or \(\arctan(1)=x\)

Now, \(\arctan(0)=0\)

and \(\arctan(1)=\dfrac{\pi}{4}\)

But the tangent function is periodic over \(n \pi, ~~n\in \mathbb{Z}\)
So you should actually have:
\(3x=0+n\pi\)
\(x=\dfrac{\pi}{4}+n\pi\)

Find all of these values for n=0,1,2,.. etc. such that you stay within the given interval \([0, 2\pi)\)

Answer 2

Thank you.

Answer 3

:)