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# solving cos 5t + 1 = 0 etc. P6 Q watch

1. You have the equation

16x^5 – 20x^3 + 5x + 1 = 0

and you know that cos 5θ = 16cos^5 θ – 20cos^3 θ + 5 cos θ

so substituting cos θ = x

cos 5 θ + 1 = 0
cos 5 θ = -1
5 θ = 180, 360, 540
θ = 36, 72, 108

arccos x = θ

so arccos x = 36, 72, 108
so x = 0.809, 0.302, -0.309

but they say that x = -1, -0.309, 0.809

Where have I gone wrong?
2. You have your calculator in degrees. Radians are said to be the natural system to measure angles with. Degrees are totally arbitrary.
3. they used degrees in the answer scheme. thats the only reason i did. the first time i did the question i used radians.
4. (Original post by lesser weevil)
they used degrees in the answer scheme. thats the only reason i did. the first time i did the question i used radians.
Sorry, I can see now that it doesn't matter as the angle is only the intermediate. I think the problem is that 5θ360.
5. (Original post by lesser weevil)
You have the equation

16x^5 – 20x^3 + 5x + 1 = 0

and you know that cos 5θ = 16cos^5 θ – 20cos^3 θ + 5 cos θ

so substituting cos θ = x

cos 5 θ + 1 = 0
cos 5 θ = -1
5 θ = 180, 360, 540
θ = 36, 72, 108

arccos x = θ

so arccos x = 36, 72, 108
so x = 0.809, 0.302, -0.309

but they say that x = -1, -0.309, 0.809

Where have I gone wrong?
cos(360) is 1
6. Yep, 5 theta= -180, 180, 540
7. ah. but how do I know that I should use a value of theta that is negative? .....
8. (Original post by lesser weevil)
ah. but how do I know that I should use a value of theta that is negative? .....
cos(-180/5)=cos(180/5) so that doesn't really matter.
You can use 5θ=180,180+360,180+720,180+1080. .. until you get to θ>360.
I assume you'll get θ=pi eventually.
9. it gives as an answer x = -1. How do I get to that value?
10. (Original post by lesser weevil)
it gives as an answer x = -1. How do I get to that value?
5θ=180,540,900
θ=36,108,180
x=..,..,-1.
11. ooh i see... thanks gaz...

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