Solving sin^2x+cos^3x=7/8

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I am a Math teacher and it is awesome to observe that, there are angles which we can be used for solving a cubic equation. That's literally awesome. :) For example: it is generally unknown, that cosinus(pí/5) = 36 degree is 1+sqrt(5)/4. But if we know this, than we can solve a cubic equation, it is simply amazing. :)

peterkrenn
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Him saying "stay safe" is a constant no one talks about.

naradmaxwell
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If you are not interested in specific values of x and only in a general solution, then i prefer to write the solution to cos(x)=1/2 => x=+-(π/3)+2πn and also for the other solutions.
It's much shorter and "cleaner".

udic
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Hey! Really enjoyed this one, especially finding the second pair of solutions. One thought: would you consider reversing the contrast on your graphs to match the dark black background? I watch the previous day’s videos on the treadmill early in the morning in a mostly dark house and the switch from dark to blindingly white is always a bit jarring. Regardless: I love the videos! They’re a perfect warm-up for a day teaching math. Thank you!

mrmikejsteele
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you can make the T=pi for the cosx=cos(5/3pi)

krnyvgy
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Wow, your problem choices are great!
I try to solve it before seeing your explanation. Sometimes there is success but sometimes I need to follow you.
Thanks for giving us brain food

mathplus
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I find these types of problems so interesting because of the neat and tidy ways you can make polynomial expressions, then abstract on the trigonometric functions, turning them into some variable. (y=cos(x), y^2=cos^2(x)). It's one of those cute instances where the annotation of mathematics betrays the actual meaning of what you're trying to express.

It probably would have been prudent of me to try and factor the polynomial instead of going immediately to Newtonian iteration, but, hey, I have this calculator for a reason.

xleph
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the expression is even ( f(x) = f(-x)), just study it for x >=0 and complete with the symmetrical

WahranRai
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Nice, I did the same before to watch the video. I loved the Golden flavor 😍 @SyberMath, may I suggest a video from another channel for who are more interested about understanding pi/5 AND phi relationship?

Drk
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Thanks for your imaginative sol'n. I think though, here, (as eg suggested by X J Will below) it is not difficult (even for me!) to spot that if 'f(p) = 'p^3 - p^2 + 1/8 = 0', where' p = cos (x)', then 'f(1/2) = 0'. Maybe, we could also just put 'q = 1/p' ( I know you like substitution!), to give 'q^3 - 8q + 8 = 0', leading to 'q = 2', and 'q = + or - (5)^(1/2) + 1', etc.

timc
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I felt it difficult, but you made it simple.Thanks.

shrikrishnagokhale
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Although, to tell the truth this method is very familiar, because I counted out cosinus(72) degree 2 weeks ago from a Geometric problem and I was very happy. But it is a really good feeling, that I was thinking well.

peterkrenn
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I tought that we should solve it with delta, and you know that its too hard, but always we have a easier method
Thank you bro❤

hadiamoohashemi
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Let c = cos(x), s = sin(x)
s^2 + c^3 - 7/8 = 0
1 - c^2 + c^3 - 7/8 = 0
8*c^3 - 8*c^2 + 1 = 0
(2*c - 1) * (4*c^2 - 2*c - 1) = 0
c = 1/2, ( 1 ± sqrt(5) ) / 4
arccos left as exercise for the reader

XJWill
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What if you let c=cos^2(x) and get a quadratic instead of a cubic?

maxhenderson
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i like method 3, because all you have to remember is pythagoras

davidseed
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My guess before watching is to turn sin^2 into 1 - cos^2 and solve a cubic equation

P-
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Acceptable as long as you away from geuss and check

abdoshaat
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What is the name of the program that made the graph? thanks.

deocleciomotajr.
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From inspection we see cos x = 1/2 satisfies the equation

kinshuksinghania