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[tt_deeb]

Hey guys, I'm having a little trouble finding out the answers to these, can someone show me how to do it.

Got to solve for exact answers from 0 < x < 2pi

I think I almost got the first one:

1.) sin(x)+1= 2cos(x)^2

(1-cos(x)) + 1 = 2 cos (x)^2

2 = 2 cos(x)^2 + cos (x)

2 cos(x)^2 + cos (x) - 2

2 (cos(x)^2 + 1/2 cos(x) - 1)

(can't really factor properly so I assume I did something wrong along the way)


2.) 2csc(x)^2=cot(x)+3

Thanks for any help.

 

[Aruarian Reflection]

This is just testing your application of trig properties, right? Pull out a trig table and see if that helps. I would actually solve the problems, but I'm leaving for the Duke vs. Georgia Tech game in 10 minutes.

 

[tt_deeb]

This is just testing your application of trig properties, right? Pull out a trig table and see if that helps. I would actually solve the problems, but I'm leaving for the Duke vs. Georgia Tech game in 10 minutes.

Are you sure? I don't think a trig table would help me out. Initially, I have to (not even sure what its called) factor out the equations so I can get something easy like Cos(x) = 1.

This is where I'm stuck on the second one:

2.) 2csc(x)^2=cot(x)+3

2 (1/sin(x))^2 = (cos(x)/sin(x)) + 3

 

[Memles]

Yeah, you're a little off on the first one.


1.) sin(x)+1= 2cos(x)^2

(1-cos(x)) + 1 = 2 cos (x)^2 (You can't do that. The property is 1 = cos(x)^2 + sin(x)^2, that's how you have to do it.

So...

sin(x) - 1 = 2(1 - sin(x)^2))
-sin(x) = 2sin(x)^2 - 1
0 = 2sin(x)^2 + sin(x) - 1

(2sin(x) - 1)(sin(x) + 1) = 0

2sin(x) - 1 = 0 and sin(x) + 1 = 0
sin(x) = 1/2 sinx = -1
x = Pi/6, 5Pi/6 (I think) x = 3Pi/2

And, while my High School Calculus teacher may murder me for this (He'd know, somehow), as an English major I completely forget the Cosecant and Cotangent properties, but they aren't too hard.

 

[bachikarn]

For the second one:

Ok you know sin^2 + cos^2 = 1
If you divide everything by cos^2 you get:
tan^2 + 1 = sec^2
If you divide sin^2 + cos^2 = 1 by sin^2 you get:
1 + cot^2 = csc^2
So you can use that to solve the second one

2csc(x)^2=cot(x)+3
2(1+cot^2) = cot +3
2cot^2 - cot -1 = 0
(2cot +1)*(cot-1) =0
cot = -1/2 and cot = 1

so x = Arccot(-1/2) [from 0 to 2pi], pi/4, 5pi/4

 

[BooM235689]

those are so easy Idiot!!!

 

[pestul]

those are so easy Idiot!!!

"What's todays letter kids?"
"Beeeeeee"
"And what's a word that starts with the letter 'B'?"
"BANNED"

 

[Dilbert]

those are so easy Idiot!!!

Either post a solution or a hint, or shut the fuck up.


(pestul, be warned that some other mods/admins are banning...for recommending bans.)

 

[tt_deeb]

Thanks guys!

 

[BooM235689]

i am so sorry for the misunderstanding i was kidding, u see i know tt_deeb infact i live next to him i was only playing (he will denie it but it is true). Look ill even complement him tt_ddeb you are amazing at tetris attack.

 

[BooM235689]

*bump*