Oh yeah! i see now!
it will come out to cosx=-1/2 and 1 which is "120°" and "0°" which is why it has to be 0°+120k°.
It was just some simple wrong factoring that got me on the wrong track.
Thanks!
So that means i mest up somewhere in the factoring? and its not suppose to both equal 1?
Well, i get this after i factor 2cos²x-cosx-1=0 :
cosx(2cos-1)-1=0
I am really stuck on this problem.
ok here's a question i can't figure out why the answer is the way it is.
Solve each equation for all values of x.
1.cos2x=cosx
*First i use the double identity 2cos²x-1 for cos2x and got:
2cos²x-1=cosx
*then i subtracted cosx and got:
2cos²x-cosx-1=0
*then i subtract 1 on both...
OOPS! i guest i made a lot of mistakes.:frown:
ill go back over my work and reread the chapter, and thanks for checking my answers.
oh yeah, and it was 0≤x≤180...... i am really sorry i mistyped it.:frown:
I have a trig problem that i need help with. I have done a) & b) but not c) because i don't get it.
Solve the following for:
a) principle values of x
b) 0°≤x≥180°
c) all values of x
1. 2sin²x-1=0
my work:
1a. principle values of x
sinx(2sin)-1 = 0...
Ok, here's the deal, this is a physics problem that might seem basic to you. But to me its a pain, I've been working on this problem for 2 days now, and my group members are of no assistance to me. They have different answers than i do which i didn't agree to.
The question
Jessica runs up a...