Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- rational

solve \(\sin^5x + \cos^5x=1\)
\(x\) is in first quadrant

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- rational

solve \(\sin^5x + \cos^5x=1\)
\(x\) is in first quadrant

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- rational

Easy to see that \(x=0,\pi/2\) are trivial solutions
and wolfram gives
http://www.wolframalpha.com/input/?i=solve+sin%5E5x%2Bcos%5E5x%3D1%2C+0%3Cx%3Cpi%2F2
idk how to prove that solutions don't exist for \(0\lt x\lt \pi/2\) ..

- anonymous

looks cool
http://www.wolframalpha.com/input/?i=x^2%2By^2%3D1%2Cx^5%2By^5%3D1

- rational

that looks cool indeed xD
i know it boils down to \(\sin^2x+\cos^2x=1\) somehow

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

that is your only real condition right?

- rational

if you mean x is real, yes..

- anonymous

you are solving \[x^5+y^5=1\] given that \[x^2+y^2=1\]

- rational

Ahh right! both problems are equivalent

- rational

and i feel equally hard/easy

- anonymous

yeah and i wouldn't bet more than $7 that there is not some more snappy approach

- anonymous

maybe just graph \(x^5+y^5=1\)

- rational

I think so... is it sufficient if we could show that \(x^5+y^5\lt x^2+y^2\) for \(x,y\in (0,1)\)

- rational

Wow! that does it! thnks @satellite73

- rational

sure calculus can solve everything :)
it is easy to notice that \(a^n \lt a^m\) for \(a\in (0, 1)\) and \(n\gt m\)

- rational

its not stupid, i think its an alternative approach...

- rational

Here is the complete proof :
for \(x\in(0,1)\) we have \(\sin^5x\lt \sin^2x\) and \(\cos^5x\lt \cos^2x\).
that implies \(\sin^5x+\cos^5x \lt \sin^2x+\cos^2x=1\) \(\blacksquare\)

- ikram002p

that solves it

Looking for something else?

Not the answer you are looking for? Search for more explanations.