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# Here we go again trig mk6 Watch

1. This time -180°≤x≤180°

cosx(cosx-2)=1
expands to cos²x-2cosx-1=0
Let cosx=y
y²-2y-1=0

using the quadratic formula gets me -2±Square root of 4+4 all divided by 2

cosx=0.41
so x is 65.5°
why have i gotten the wrong answer?
2. Off the bat shouldn't it be (+2±√4+4)/2
3. y^2-2y-1=0 -> (y-1)^2-2=0 -> y-1=+/-sqrt(2) -> y=1+/-sqrt(2), and since -1<=cosx<=1, cosx=1-sqrt(2)=minus 0.414...
Thus cosx=1-sqrt(2) -> x=114.5, 245.5.
4. (Original post by GUMI)
Off the bat shouldn't it be (+2±√4+4)/2
nooooooooooooooooooooooooooooooo oooooooooooooooooooooooooooooooo oooooooooooooooooooooooooo my Fateful mistake of signs noooooooooooo i never though this would happen, my life is over...
Spoiler:
Show
Cheers
5. (Original post by thefatone)
nooooooooooooooooooooooooooooooo oooooooooooooooooooooooooooooooo oooooooooooooooooooooooooo my Fateful mistake of signs noooooooooooo i never though this would happen, my life is over...
Spoiler:
Show
Cheers
haha np m8
6. (Original post by constellarknight)
y^2-2y-1=0 -> (y-1)^2-2=0 -> y-1=+/-sqrt(2) -> y=1+/-sqrt(2), and since -1<=cosx<=1, cosx=1-sqrt(2)=minus 0.414...
Thus cosx=1-sqrt(2) -> x=114.5, 245.5.
Yup cheers for pointing the signs thing out also x has to be between -180° and 180°
7. Yeah ok x=+/-114.5

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