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Help! Trignonometry

Hi, I need some quick help with two questions:

1) 4(sin^2X - cosX) = 3 - 2cosX (quadratic formula?)

2) cos^2 3X - cos3X = 2

Cheers!
Reply 1
Avatar for TenOfThem
TenOfThem
18
The second one is already a quadratic ( in Cos3x)

The first one needs you to use sin2x+cos2x=1sin^2x+cos^2x = 1 then that will be a quadratic too (in Cosx)
Reply 2
Avatar for I love Vagina
I love Vagina
OP
Original post by TenOfThem
The second one is already a quadratic ( in Cos3x)

The first one needs you to use sin2x+cos2x=1sin^2x+cos^2x = 1 then that will be a quadratic too (in Cosx)


Would you mind showing me how to do the second question? :smile:
Reply 3
Avatar for TenOfThem
TenOfThem
18
Original post by I love Vagina
Would you mind showing me how to do the second question? :smile:


Can you solve

x2−x−2=0x^2 - x - 2 = 0
Reply 4
Avatar for I love Vagina
I love Vagina
OP
Original post by TenOfThem
Can you solve

x2−x−2=0x^2 - x - 2 = 0


So cosX = -1 or 2 (however we ignore the 2).

And this is where I'm stuck.. it wants it within the rane of -pi and pi.
So would I need to change the range to solve?
Reply 5
Avatar for TenOfThem
TenOfThem
18
As you say Cos3x = -1

so what can 3x equal

you need a bigger range for 3x but not for x
Reply 6
Avatar for I love Vagina
I love Vagina
OP
Original post by TenOfThem
As you say Cos3x = -1

so what can 3x equal

you need a bigger range for 3x but not for x


So..
cos 3x = -1
3x = pi
Hence x = pi/3 ??
Reply 7
Avatar for TenOfThem
TenOfThem
18
Original post by I love Vagina
So..
cos 3x = -1
3x = pi
Hence x = pi/3 ??


and
Reply 8
Avatar for I love Vagina
I love Vagina
OP
Original post by TenOfThem
and


-pi/3?
Reply 9
Avatar for TenOfThem
TenOfThem
18
Original post by I love Vagina
-pi/3?


Cos3x = -1

gives infinite answers

More than 2 of these will give x in your range
Reply 10
Avatar for benten17
benten17
Ohh. I did all of this a few months ago and now I don't remember a thing! That's the bad thing about summer :frown:

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