Question

If you solve a trig equation and you get the solutions shown below, find all solutions on the interval [-2pi, 2pi)

Answer 1

\[[\frac{ \pi}{ 4 }+2\pi n, \frac{ \pi }{ 3 }+ 2\pi n, n \in \mathbb{Z}^+ ]\]

Answer 2

that will give the solutions between [0,2pi] if you set n=0
that is consider the first rotation

if you had wanted the solutions in [2pi,4pi] you set n=1

can you make a guess as to what you what set n to in order to get the solutions in [-2pi,0]?

Answer 3

n= -1 ?

Answer 4

yes

Answer 5

so to get the solutions in [-2pi,0] set n=-1
to get the solutions in [0,2pi] set n=0
this will give you all the solutions in [-2pi,2pi]

Answer 6

ok so use the solutions given above and set n to -1 and 0 to find the other solutions?

Answer 7

you will replace all the n's with -1


then you will replace all the n's with 0

Answer 8

the weird thing is Z^+ means positive integers only

Answer 9

So that is odd
and weird since to find the solutions in [-2pi,2pi]
we need non positive n's

Answer 10

o ok, it was supposed to be all all solutions, not only positive. my bad

Answer 11

anyways, thanks for showing me the way how to do it, bud. I was confused, reading the question and was not sure what to do.

Answer 12

np