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PartyBullshiit
March, 2024

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Need help with trig... (+k)

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7springsprod
Posts: 2661 - Karma: 3,181
How do I do: 8 cos2x + 3 = 2 cos2x- 1. Ive tried using identities to change cosine to sine and I still have no clue what to do. +K for help.
Mar 21 2012 4:44PM
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*DELETED*
Posts: 1043 - Karma: 938
solving for x? or just simplifying?
Mar 21 2012 4:49PM
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7springsprod
Posts: 2661 - Karma: 3,181
I can solve them, but Im struggling at simplifying this onr because usually they are quadratics.
Mar 21 2012 4:53PM
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roddy116
Posts: 10622 - Karma: 7,397
try to get it all to one side then put it in a calculator?
Mar 21 2012 4:53PM
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Radio
Posts: 946 - Karma: 12
you have not taken trig
Mar 21 2012 4:54PM
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Broseph_Stalin
Posts: 3745 - Karma: 177
alright so you do 8 cos2x- 2cos2x equals -4

then 6 cos2x = -4

then cos2x = (-4/6)

then cosx = (-4/6)^(1/2)

then x = arccos((-4/6)^(1/2))


i did this in like 2 seconds so i might be wrong someone check me
Mar 21 2012 4:54PM
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Broseph_Stalin
Posts: 3745 - Karma: 177
and if thes a cos(2x) instead of a cosine squared x (like i did above hopefully correctly) then..

you do 6cos(2x) = -4

then cos(2x) = (-4/6)

then 2x = arccos(-4/6)

then x = (arccos(-4/6))/2
Mar 21 2012 4:59PM
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Alex.W
Posts: 528 - Karma: 22
Got beat to it by BrosephStalin.

You can't use your sin^2x+cos^2x=1 identity because your trig functions aren't squared btw. Subtract the 2 functions to get one and solve from there.
Mar 21 2012 5:07PM
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