trig equation help

[grave harness]

Basically like the first question but you gotta find solutions between

$\ -3\pi \leq x < -\pi$

[muted girder]

adonhs

[devout pasture]

Between that its 2pi right ?

-2pi

[grave harness]

-2pi is in that intervall correct

Find all solutions between -3pi and -pi

[devout pasture]

Idk how to find it

[grave harness]

How did you find the solutions then in the first one xD

Anyway let's do it together:

$\ \frac{4\pi}{3} + 2\pi n$ ...first solution set

$\ \frac{5\pi}{3} + 2\pi n$ ...second solution set

[muted girder]

adonhs

[devout pasture]

What i did on the first one

[grave harness]

Ok let's start with the first set

we need to go into the negative

so we subtract n multiples of 2pi from 4pi/3

are you allowed to use calculator?

[devout pasture]

No

[grave harness]

ok no problem

we can rewrite $2\pi = \frac{6\pi}{3}$

[muted girder]

adonhs

[grave harness]

Now let's subtract this from 4pi/3

we get -2pi/3

is this value between -3pi and pi?

[devout pasture]

Yes

[grave harness]

there you go first solutions now we go further

we subtract another

-2pi/3 - 6pi/3

-8pi/3

[devout pasture]

But they got -8pi over 3

Hold up im confused

[devout pasture]

Why am I subtracting it twice?

[grave harness]

because each solutions 2pi away

[devout pasture]

That makes sense!

[grave harness]

we can estimate -8pi/3

-8pi/3 > -8*3/3 = -8

-8 is this value in -3pi and -pi

[devout pasture]

if it was -3 pi way, I’d subtract it three times?

[grave harness]

hold on

this is what we doing right now

we periodically go throuch each solution by adding/subtracting one period

[devout pasture]

Ok I understand now, thank you!

[grave harness]

you can imagine it like this

we got some area (intervall technically)

and we are looking for solutions there

now we start in 4pi/3 somewhere positive so we have to go back

and we do this by subtracting always multiples of the period since each solutions is multiples of the period away from each other

So we got our solutions:

$\x_1 = -\frac{8\pi}{3}$

$\x_2 = -\frac{7\pi}{3}$

if we go further then those solutions will be out of the intervall

wait no

-2pi/3 is less than pi

[devout pasture]

The second solution is -7pi/3

[grave harness]

so it's not betwee -3pi and -pi

[grave harness]

you did the same with 5pi/3 by subtractin one periode, right?

[devout pasture]

Yea

[grave harness]

well done

so we have actually 2 solutions in that intervall

i just realized that's actually the solution you provided up there😂

[muted girder]

adonhs

[devout pasture]

I appreciate the explanation tho, thx

[grave harness]

If there's no questions you may close the case with .solved

[devout pasture]

.solved