Further trig question (1 Viewer)

[hs17]

Hi can somebody please help me with this question

 

[Etho_x]

Hi can somebody please help me with this question

View attachment 29025

What do you have to do

 

[Greninja340]

just solve for tan3x-tanx=0

 

[hs17]

Oh sorry haha

"solve for the given domain..". So solve for theta

 

[Etho_x]

Oh sorry haha

"solve for the given domain..". So solve for theta

I’ll get onto it now no worries

 

[Etho_x]

I’ll get onto it now no worries

Hopefully you can see this, I did it on my whiteboard lol (not sure if this is correct and I’m sure someone could provide a better solution)

 

[hs17]

Those are the solutions I got as well but the answers say
theta = 0, pi/2, pi, 3pi/2, 2pi

Thanks btw

 

[B1andB2]

Hopefully you can see this, I did it on my whiteboard lol (not sure if this is correct and I’m sure someone could provide a better solution)

one question, why does the domain stay the same for tan2phi, and not become 0<= 2phi<=4pi

 

[Etho_x]

Those are the solutions I got as well but the answers say
theta = 0, pi/2, pi, 3pi/2, 2pi

Thanks btw

But I wouldn't think it'd be pi/2 or 3pi/2 because it's undefined for both tan3x and tanx

 

[Etho_x]

one question, why does the domain stay the same for tan2phi, and not become 0<= 2phi<=4pi

Because dividing through by 2 and solving for theta would give you the same domain, or perhaps I might be wrong on that but regardless if the answers are between 0 and 2pi i could only imagine that's what the answers are

 

[B1andB2]

Because dividing through by 2 and solving for theta would give you the same domain, or perhaps I might be wrong on that but regardless if the answers are between 0 and 2pi i could only imagine that's what the answers are

hmmm, if you change the domain you get 2phi=3pi as a possibility then divide by 2 : phi = 3pi/2 which is one of the answers... (according to hs17)

 

[Etho_x]

hmmm, if you change the domain you get tan2phi=3pi then divide by 2 : phi = 3pi/2 which is one of the answers...

Hmm you might be correct I'll just double check with my working, but that doesn't make sense anyway because it'd be undefined

 

[B1andB2]

Hmm you might be correct I'll just double check with my working, but that doesn't make sense anyway because it'd be undefined

yeah, tan(270 x 3) = undefined = tan(270) = undefined so it satisfies the q

 

[Etho_x]

yeah, tan(270 x 3) = undefined = tan(270) = undefined so it satisfies the q

Ah yep with the domain I see where I went wrong, love careless mistakes! However with that being said unless they're counting undefined solutions only 0, pi, and 2pi are real and not undefined solutions

 

[hs17]

wait can someone explain to me the domain situation ahaha?

 

[Etho_x]

wait can someone explain to me the domain situation ahaha?

Oh because you have a double angle the domain changes and doubles (ignore what I wrote about the domain on the whiteboard). I haven't done one of those questions in a little so I needed to jog my memory a bit hahaha.

 

[B1andB2]

wait can someone explain to me the domain situation ahaha?

Since you have tan(2phi) instead of just tan(phi) you need to change the domain to cater for 2phi. The domain in the q is just phi so times it by 2 to get 2 phi between 0 and 4pi.

This gives you more solutions

So then you're left with 2phi = blah blah

Divide all the values by 2 to get phi.

Sorry this is probably a very bad explanation im in the car lol

Basically, whatever is in the tan bracket needs to be what is in the domain

 

[Etho_x]

Ye what she said ^

 

[hs17]

Okay, thank youuu both of you!!

 

[vernburn]

Another solution could be: is periodic. So, and must differ by some multiple of . i.e.



for the given domain, let :
. This is where I believe the answers in the textbook got the extra solutions of and from.
However, you are correct to exclude them because tan is undefined for those values.
Therefore the answer should be: