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sorry I tried, but it doesn't seem straight forward.

its okay, at least you put effort into it

This isn't as hard as it looks :)

Do you need it to be explained more?

yes if you don't care to

Is there a specific part?

You can refer to the part by the number on the attachment, if that's easier.

brb

yes that made it alot easier to see thank you! do u care to help me on another one?

Oh, so you're okay with the last one?

Good. What's your other question?

solve this trig equation on the interval 0

2 or 3 more and we can start a club ;)

It can be factored into \[(2u+1)(u+1)=0\]

Yeah, but in 30s you'll be on infinity hero, so... :)

lol.... by then ill just need another account :)

No, by then it's like TRON and you're sucked into the website.

Sorry mj

is it (2cos^2+1)(cosx+1) btw no harm here i aprreciate your help!

Yes it is.

That whole thing is equal to zero, though.

how do i find the angles? thats where i have a problem

Yep...I'll do them on paper first.

Okay thanks!

The square of any real number is always 0 or positive.

If you plu x=pi into your original equation, you will get 0.

*plug

Thanks! if i had (sin(x)-4)(sin(x)+1) how would i find the angles on the same interval?

Is that equal to zero?

yes sorry i forgot

*'Sot' = 'For'.

is it -pi/2 +2kpi =3pi/2

Yes it is.

3pi/2

so sinx=3pi/2

No, \[\sin \frac{3\pi}{2}=-1\]

You're trying to find the x's that, when plugged into sine, give -1.

oh so x=3pi/2 i made a mistake

Yes.

you have been a lifesaver, that helped me out alot, especially with the identities.

You're welcome, mj :)

So that's it? You're done?

Uhm this one seems simple but i forgot the identity to use
5sin^2x+2cos^2x=5-3cos^2x

The statement is either true or false. If it's not false, it must me true.

If you're not happy with that way, I have another.

Yeah...I'll do it another way.

ahh thats exactly what i was looking for i was thinking you would use sin^2+cos^2=1 for this

No more questions then?

tan(x/2)sinx=1-cosx
LHS
tan(x/2)sinx=1-cos/sin(sin/1)=1-cosx
did i do that right?

I'm not sure what you did. This is what I would do:

That was a much easier way of doing it, identities must take alot of practice to get use to.

Yeah. It's mostly intuition as to how to start and how to close. That comes from practice!

I can tell the preparation and follow through is the hardest part!
cos(x/2)
cos(x/2)=square root of 1+4/5/2
=square root of 9/5/2 =square root of 9/10
is that right if cosx=4/5 with 3pi/2

Um, it's hard for me to tell since I'm not sure of your order of operations.

What's under the square root?

\[\frac{\sqrt{1+4/5}}{2}\]?

i used the half angle formula for cos YES that is correct

Yes, and it's positive root if you're in that interval.

Do you know how to use the equation editor? The button below will take you there.

oh i didnt know that thank you!

\[\tan2(\theta)=1-\cos(2\theta)\div2\div1+\cos(2\theta)\div2 \]

yes thats what i was trying to do i got -3/13 for the answer but im not sure about it

Are you trying to find tan(2 theta) and the right-hand side is what you came up with?

yes im trin to find tan(2theta) on the interval 3pi/2

yes thats what i have it equal to

ok...let me see.

okay!

okay i will do that!
can you tell me how to simplify real quick
(-1/2)(4/5)-square root of 3/2(3/5)

I need you to use the equation editor ::

\[-1/2(4/5)-(\sqrt{3/2})(3/5)\]

\[-\frac{1}{2}\frac{4}{5}-\frac{3}{5}\sqrt{\frac{3}{2}}\]do you mean this?

yess that is right

Thank you for everything! I appreciate your time!

welcome

did you finish the last one?

nope :/

Is this due online or something?

no i have to turn in next week

Alright sounds good! I'll be sure to check back on here for it! Have a good day kind sir

You too ;)

Here you are...remember to check it.

The plot is for
f(t)=t^3-t+2