Help! Trignonometry Watch

I love Vagina
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#1
Report Thread starter 6 years ago
#1
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!
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TenOfThem
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#2
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The second one is already a quadratic ( in Cos3x)

The first one needs you to use sin^2x+cos^2x = 1 then that will be a quadratic too (in Cosx)
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I love Vagina
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#3
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(Original post by TenOfThem)
The second one is already a quadratic ( in Cos3x)

The first one needs you to use sin^2x+cos^2x = 1 then that will be a quadratic too (in Cosx)
Would you mind showing me how to do the second question?
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TenOfThem
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(Original post by I love Vagina)
Would you mind showing me how to do the second question?
Can you solve

x^2 - x - 2 = 0
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I love Vagina
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#5
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(Original post by TenOfThem)
Can you solve

x^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?
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TenOfThem
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#6
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As you say Cos3x = -1

so what can 3x equal

you need a bigger range for 3x but not for x
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I love Vagina
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(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 ??
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TenOfThem
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(Original post by I love Vagina)
So..
cos 3x = -1
3x = pi
Hence x = pi/3 ??
and
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I love Vagina
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#9
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(Original post by TenOfThem)
and
-pi/3?
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TenOfThem
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#10
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(Original post by I love Vagina)
-pi/3?
Cos3x = -1

gives infinite answers

More than 2 of these will give x in your range
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benten17
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#11
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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
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