A Rare Quintic Trigonometric Equation | sin^5x+cos^5x=1

[@syberrus]

There is a much easier way. Given that sin^2 + cos^2 = 1, and 0

[@SyberMath]

Cool!

[@playomar5106]

Wow.

[@SanctBlack]

On 10th second I though about the same solution when imagine picture of c^5+s^5

[@garvellokenxvi]

0

[@RigoVids]

The only thing to note which maybe falsely concluded from this explaination is the range of cos^5(x). The range or codomain of the function is -1

[@SyberMath]

I don't think this problem has been solved and published on the internet before. Correct me if I'm wrong. I'm kind of surprised why it wasn't because it is not a super original idea. Anyone can come up with an equation like this. The fun part was to test if one equation would imply the other. I knew that one way implication worked but was not sure about the if-and-only-if scenario! Generally quintics are not solvable. (Isn't that sad?) Well, I think they are awesome! Any thoughts? Plz share down below...

[@yogamulyadi9183]

I ever seen this on your Twitter.., And thanks to that, I learned several things...

[@serands1840]

S² + C² = 1. S^5 = S² iff S = 0 or 1. Idem for C. Otherwise S^5 < S² and C^5 < C². (S and C between -1 and 1) Thus S^5 + C^5 always < 1. Thus the only solutions are for (S,C)=(0,1) or (1,0)… i.e. 0 and pi/2 mod 2pi…

[@SyberMath]

Good thinking!

[@BCS-IshtiyakAhmadKhan]

I also did the same way

[@davidseed2939]

This method is used on youtube for C^100+S^100=1 as a competition question

[@dustinbachstein3729]

Yeah, this video could have been made 22 minutes shorter xD

[@TJStellmach]

S^5+C^5

[@hosseinmohammadi2800]

The good point about your way is that we can find the imaginary answers easier. I liked it

[@kingbeauregard]

I took the derivative of the function to see where the extrema were, and found out that the maxima were at x = 0 or pi/2, which also happened to be where the function equals 1.

[@SyberMath]

Nice!

[@VSN1001]

I did the same and found out that there are infinity many sol

[@telnobynoyator_6183]

@@VSN1001 yeah bc cos(x + 2pi) = cos(x) and so on

[@bilal42nga47]

.

[@cheesefrogsnail]

Yes I did the same, there is another local maximum in 5PI/4 but it is negative

[@bkclaud2135]

Thinking out of the box👏🏾. You literally had to come up with another equation and solve by substitution with the original problem. Wow ok

[@SyberMath]

Thanks! 😊

[@marklevin3236]

Much simpler than that for any x sin^5(x)

[@phandinhthanh2295]

I don't really get it. Could u, please, elaborate?

[@italixgaming915]

@@phandinhthanh2295 Easier to see in my solution: cos^5(x)+sin^5(x)=cos²(x).cos^3(x)+sin²(x).sin^3(x). 1=cos²(x)+sin²(x). Then I can write your equation under the form: cos²(x).cos^3(x)+sin²(x).sin^3(x)=cos²(x)+sin²(x) cos²(x).[cos^3(x)-1]+sin²(x).[sin^3(x)-1]=0 But cos^3(x)-1

[@luismuller6505]

21:35 "If you divide both sides by 0" is the best thing in the derivation of the answers. 😂

[@SyberMath]

😁

[@alax155]

21:37 "And if you divide both sides by 0". At first, I was like "Wait, whaaaaat", but then I saw you meant "by 2" xD

[@benvickers8821]

What a fantastic explanation of the solution to this problem. Love your videos!!!

[@SyberMath]

Glad you like them!

[@avrahambercovich667]

@@SyberMath אני חושב אני אוהב א אני את זה אבל אמא א

[@fm7490]

I solved for tan(x/2) For substitution: cosx= (1-t^2)/(1+t^2) Sinx=2t/(1-t^2) Where t=tan(x/2) Then find t, then x/2, then x

[@SyberMath]

Interesting

[@AtomicMathematics]

This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution kzbin.info/www/bejne/rZutfJuFj56SgZo

[@vascomanteigas9433]

If you apply the Weierstrass substitution: sin(x)=2*t/(1+t^2) and cos(x)=(1-t^2)/(1+t^2). It will get a 10th degree polynomial. Each Root are then derived from x=2*arctan(t)+2*pi*n.

[@SyberMath]

Wow!

[@זאבגלברד]

To solve sinx + cosx = 1 there is a known technic to write it as sinx + tg45 cosx =1 , then multiply both sides by cos45. You get sin(x+45)=cos45 ...

[@SyberMath]

That's a good one!

[@akshaybagaria4941]

Solving this with inequality x^2>x^5 when x

[@meowmeow-yq9xt]

Thanks I like this problem You guys are one of my hope that i will learn new and it needed from the beginning but my country education system doesn't fit in me. I can't even do something . But when I got my phone I realized that I shouldn't misuse of this device then step by step I found you and others who teaches free in YT at first I searched for graphical dynamic , optics , mechanics and other physic related then your one video came as recommendation I watched and I liked the way you teach and I can clearly understand your accent then it made me to subscribe your channel . Thanks for the videos

[@זאבגלברד]

I enjoy your math telent. Thank you !

[@SyberMath]

You are welcome! Thank you!

[@sekaibelle]

"Hello everyone" :)

[@SyberMath]

Seeing this comment again after 2 months! 😁

[@tonywong8677]

Solved by plotting graph. The zeros of the function are at 0 and n pi/2. Hence, we can find out if there is any other solutions by plotting the graph of the corresponding function.

[@AtomicMathematics]

This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution kzbin.info/www/bejne/rZutfJuFj56SgZo

[@Mosux2007]

The quintic can also be factored using synthetic division. May be easier.

[@krisbrandenberger544]

On the one part, he said "Divide by 0", but meant "Divide by 2."

[@hellohabibi1]

21:36 "And If you divide both sides by 0"😅

[@eliasmazhukin2009]

In Russia, we call "sin^2 (x) + cos^2 (x) = 1" the Fundamental Identity of Trigonometry (основное тригонометрическое тождество)

[@AtomicMathematics]

This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution kzbin.info/www/bejne/rZutfJuFj56SgZo

[@peterwan9076]

@@AtomicMathematics I watch your solution. I appreciate the part where you figured out S^2(1 - S^3) + C^2(1-C^3) = 0 => S^2(1 - S^3)=0 and C^2(1-C^3) = 0. Genius. After that, it is straightforward.

[@omsingh7683]

As sine and cosine function are the x coordinate and y coordinate of an arbitrary point on the unit circle centre at the origin therefore sin^2x>=sin^5x and cos^2x>= cos^5x now adding these two inequalities we get sin^5x+cos^5x

[@dneary]

For the last bit, you can use the "trick" of multiplying acorss by 1/sqrt(2): sin x cos pi/4 + cos x sin pi/4 = sin(x+pi/4) = 1/sqrt(2) implies x+pi/4 = pi/4 or 3pi/4, x=0 or pi/2

[@vijaysrini27]

How did you get the cubic equation solved? Cubic equation direct problem hasn't been taught in either our school or college till date, and is still a mystery to most people, even in engineering!!!

[@SyberMath]

There's a cubic formula. I used that method many times

[@shubhamjha6418]

from which book this question had been taken?

[@SyberMath]

I don't think it was from a book. I thought of it myself

[@salloom1949]

I followed your methodology diligently and I understood every step along the way. Thoroughly enjoyed your explanation. But you left me dizzy and with headache. Thank you for a wonderful procedure.

[@SyberMath]

Excellent!

[@Jono98806]

Regarding that mistake in the end, you can write sin x + cos x = sqrt(2)*(1/sqrt(2) sin x + 1/sqrt(2) cos x) = sqrt(2)*(cos(pi/4) sin x + sin(pi/4) cos x) = sqrt(2) sin(x + pi/4). Therefore, sin x + cos x = 1 implies sqrt(2) sin(x + pi/4) = 1, where dividing both sides sqrt(2) gives sin(x + pi/4) = 1/sqrt(2). Therefore, x + pi/4 = pi/4 or x + pi/4 = (3 pi)/4 and solving for x gives x = 0 or x = pi/2.

[@grzegorzkondracki4630]

Solution like an experience during climbing on K2 in one shoe, walking backward. I can admire, but Serands and some others solved this very smart and simply.

[@prabhakarbk4322]

Too good an explanation. I enjoyed the way it is solved

[@SyberMath]

Thanks a lot 😊

[@AtomicMathematics]

This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution kzbin.info/www/bejne/rZutfJuFj56SgZo

[@miloradtomic]

Beautiful method, dear friend. Best regards from Serbia

[@SyberMath]

Many thanks! 🧡

[@ostdog9385]

Very good!!

[@SyberMath]

Thank you!

[@MayankSharma-jz9lc]

I had this question for my test and I am surprised how utube recommended this to me

[@SyberMath]

We are being tracked! 😜😂

[@jensknudsen4222]

So, all that work to show that the two solutions you can guess up front by plugging in a few points of the compass are in fact the only two solutions (mod 2pi). Isn't that just math in a nutshell?

[@tylerduncan5908]

Yes but the difference is that he said "i will find out for sure" not "I'm gonna guess and hope those are the only 2 solutions

[@dmitryweinstein315]

instead of solving the cubic equation exactly, you could have analysed the lines u^5 and 5u-4 and realised that they only have two points in common, u=1 (where the tangents of the two lines equal i.e. no other positive u) and a negative u which is less than -sqrt(2) because at u=-sqrt(2) the line u^5 is still higher than the line 5u-4

[@huytrandang277]

This, for some reason, is actually entertaining to watch.

[@SyberMath]

Glad to hear that!

[@MrArcan10]

where you have got u3 from? (this complicated value with cubic roots)

[@mikesteele5935]

This follows almost immediately from the Cauchy Schwarz inequaliity (if we restrict attention to real x, which seems to have been implicit).

[@muhandiz6585]

21:37 if you divide both side by 2* you said 0 which is was shocking. interesting solution refreshed the arithmetical part of my brain.

[@ezzatabdo5027]

Thanks for your nice work.

[@SyberMath]

Thank you too!

[@think_logically_]

sin⁵x ≤ sin²x, cos⁵ x ≤ cos²x (this also covers negative values of sine and cosine), consequently sin⁵ x+ cos⁵ x ≤ sin²x + cos²x = 1. Therefore the equality holds if and only of both inequalities turn into equalities. Thus we get sin⁵x = sin²x, cos⁵x = cos²x , so sine or cosine must be either 0 or 1. Since sine and cosine cannot be both 0, or both 1, one of them must be 1, and another must be zero.

[@宇佐見英晴]

We can use the trigonometric identity: sin^2(x) + cos^2(x) = 1 to rewrite the equation as: (sin^2(x) + cos^2(x))(sin^3(x) + cos^3(x)) = 1 Expanding the second factor using the identity: a^3 + b^3 = (a + b)(a^2 - ab + b^2) with a = sin(x) and b = cos(x), we get: (sin^2(x) + cos^2(x))(sin(x) + cos(x))(sin^2(x) - sin(x)cos(x) + cos^2(x)) = 1 Using the identity: sin(x)cos(x) = (1/2)sin(2x) we can simplify the third factor to: sin^2(x) - sin(x)cos(x) + cos^2(x) = sin^2(x) - (1/2)sin(2x) + cos^2(x) = 1 - (1/2)sin(2x) Substituting this back into the equation, we get: (sin(x) + cos(x))(1 - (1/2)sin(2x)) = 1 Expanding the first factor using the identity: a+b)^2 = a^2 + 2ab + b^2 with a = sin(x) and b = cos(x), we get: (sin(x) + cos(x))^2 = sin^2(x) + 2sin(x)cos(x) + cos^2(x) = 1 + sin(2x) Substituting this back into the equation, we get: (1 + sin(2x))(1 - (1/2)sin(2x)) = 1 Expanding and simplifying, we get: 1 - (1/2)sin^2(2x) = 1 sin^2(2x) = 0 Taking the square root of both sides, we get: sin(2x) = 0 This means that either 2x = kπ (where k is an integer) or x = (kπ)/2. Therefore, the solutions to the equation are: x = kπ/2 or x = kπ/4, where k is an integer.

[@wyboo2019]

if you allow complex valued theta then you can have those solutions outside of the regular bounds

[@user-xo1ty8dg9k]

Avant de répondre a la question / Où est la formule ?

[@Caturiya]

Anstelle der Kubischen Formel. Wir nennen das kubische Polynom K(u). Es har nur eine reele Nullstelle, da K(0) = 4 ist, ist diese negativ. etc

[@marioalejandrocaisaguanomu8005]

A really interesting problem! We can also solve it using this identity: cos^5⁡(x) + sin^5⁡(x) = [cos⁡(x) + sin⁡(x)][1 - cos⁡(x)sin⁡(x) - cos^2⁡(x)sin^2⁡(x)]. We can use "u = cos⁡(x) + sin⁡(x)". After some substitutions we'll have this equation: u^5 - 5u + 4 = 0. If we undo the substitutions, we will have: sin⁡(2x) = u^2 - 1. From the equation in terms of "u", we'll have two real solutions. We only use the "u = 1" solution because the another value of "u" makes the expression "u^2 - 1" greater than one (max(u^2 - 1) = 1). The identity that I used can be proved if we factorize "cos^5⁡(x) + sin^5⁡(x)" as "[cos⁡(x) + sin⁡(x)][cos^4(x) - cos^3(x)sin(x) + cos^2(x)sin^2⁡(x) - cos(x)sin^3(x) + sin^4(x)]" and using another basic identities.

[@kevinhermawan369]

This earth gonna be better if all lecturer like you

[@SyberMath]

Aww, thank you!!! 🧡🥰

[@momohkaddri941]

it suffices to write sin^5x + cos^5x= sin^2x + cos^2x which is eq. to this sum of potives sin^2x(1- sin^3x) + cos^2x(1- cos^3x)=0 and then, deduce , the solutions.

[@_wahahaha]

When u are obsessed with mathematics for 20mins u can even divide equations by zero ... 21:37

[@SyberMath]

😜

[@זאבגלברד]

Like you did in the (sinx)^0.5 + cosx = 0 you could say right at the begining that x is greater or equal 0 and x is less or equal pi/2 .

[@SyberMath]

That's right!

[@orenawaerenyeager]

21:37 💀witnessed a very scary statement.

[@SyberMath]

Division by 0 should be allowed!!! 😜🤣

[@orenawaerenyeager]

@@SyberMath 😂

[@mathematicsbeautyclinics7687]

Nice mathematics. Kindly tell me the app or software i need to have to also enjoy calculating that way at home

[@SyberMath]

Thanks! Notability

[@liyuan-chuanli8468]

If |\sin x| < 1 and |\cos x| < 1, then sin^n(x) is strictly decreasing on n\ge 1. Similarly, sin^n(x) is strictly decreasing on n\ge 1. So, sin^2x + cos^2x = 1 implies that sin^5x + cos^5x < 1. Therefore sin^5x + cos^5x = 1 implies that either sin x = 0 or cos x = 0.

[@CengTolga]

Nice video. Can you revisit this problem, but including complex numbers? I'm particularly interested in u₃=-1.6506

[@SyberMath]

Thanks! Me no like complex numbers! 😁

[@cdiesch7000]

interesting solution i would have done with inequalities by this we could demonstrate for sin(x)^n+cos(x)^n=1 n> 2 and natural number, there are in ]-pi,pi]2 solutions 0 and pi/2 if n is odd and 4 solutions, -pi/2,0,pi/2,pi if n is even.

[@SyberMath]

Wow! Pretty good!

[@Kris-hz1ns]

At 21:38, you say the most blasphemous thing that can be said in Mathematics !! 😆 Other than this, the whole explanation was quite gripping!👍

[@SyberMath]

I don't know what I was thinking!!! 🤣

[@Kris-hz1ns]

@@SyberMath 🤣

[@jpolanco2004]

Cos(x)^5-Cos(x)^2+Sin(x)^5-Sin(x)^2=0; from there, you can factor out 1-cos(x) and 1-sin(x), both of which give solutions when set to 0. The third factor can be reduced to 1+sin(x)cos(x)[sin(x)+cos(x)]=1+(1/sqrt(2))sin(2x)sin(x+pi/4), which is greater than 0. Hence x=0 and x=pi/2 are the only solutions, plus periodicity.

[@asfahanansari4669]

Instead of raising (Sinx + Cosx) to power 5 sin3x + Cos3x can be multiple by 1 e.g sin2x + cos2x

[@piyushdaga357]

Great video ! But I, being a class 8 student, don't know how to solve for cubic and don't use radians in trigonometry...

[@SyberMath]

I understand. Solving for the cubic is not that important. Converting radians and degrees is fairly easy.

[@Qermaq]

@@SyberMath I never learned about radians until long after I was out of school. Had to learn them with Excel as it does all its trig with radians. I agree, they are no harder than converting temperature from C to F or vice versa. I hope anyone reading this realizes they're not hard, just different from degrees, and very often more useful.

[@AtomicMathematics]

This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution kzbin.info/www/bejne/rZutfJuFj56SgZo

[@Muslim_011]

The final cubed expression have -1 as a solution too Correction I'm wrong it does not have -1 as a solution

[@WolfgangKais2]

You mean that -1 + 2 - 3 + 4 = 0?

[@Muslim_011]

@@WolfgangKais2 yes

[@WolfgangKais2]

@@Muslim_011 Well, I thought it was 2: -1 + 2 = 1 and -3 + 4 = 1, 1 + 1 = 2.

[@Muslim_011]

@@WolfgangKais2 yes you're right when I first time said that I quickly thought it is 1-2-3+4 My bad. Thank you

[@fanan2644]

What is the cubic formula used in 19:50

[@SyberMath]

I used Wolfram Alpha 😜😂

[@autf2_6]

Thx teacher very good question please dowland trigonometry questions like this really beautiful question

[@SyberMath]

Np. Thank you!

[@VerSalieri]

you could use the fact that sinx+cosx=radical2 * sin(x+pi/4)... that way you won’t be adding in solutions by squaring. Of course, that requires using the formula asinx + bcosx = radical(a^2+b^2) * sin(x + arctan(b/a)). Side note: I hate that I wasn’t taught this formula in school... and by the time i discovered it exists, I no longer needed it. As always, good content my friend. Keep up the good work.

[@SyberMath]

Thank you, my friend! 💖

[@akkyiofdolphin]

とはいえ、日本ではこの「回りくどく大変な」方法を遂行する力は養われていないように思います そのいみでとても教育的なビデオでした

[@SyberMath]

ありがとう！

[@rajakumar9377]

What value of x, which satisfy this equation ??

[@AmooBaktash]

There is an easier solution for this problem. Let f(x) = sin(x)^5 + cos(x)^5. Then, f'(x) = 5 sin(x)^5 cos(x) - 5 cos(x)^4 sin(x). After a few simplifications you get f'(x) = 5 sin(x) cos(x) ( sin(x) - cos(x) ) (1 + sin(x)cos(x) ) . Setting f'(x) = 0, you fill find that the maximum of f(x) is 1, which only happens at x = 0 , and x = pi/2 (for 0

[@ivornworrell]

How did u know to introduce u in the first place to aid solving?

[@SyberMath]

Substitution is a good method and with trig expressions like this one, it works real well

[@ivornworrell]

@@SyberMath Forgive my seeming lack of advanced Mathematical knowledge, but it seems to me that a trig problem of this nature requires previous knowledge of Trig. Identities AND knowledge in Algebraic Manipulation in order to be able to quickly "spot" which trig expression should be represented by an arbitrary letter (u in this case), to ultimately determine the solution/s.Salaam

[@albertmcchan]

sin(x)^5 + cos(x)^5 = 1 → sin(x) ≥ 0 and cos(x) ≥ 0 → u = sin(x) + cos(x) > 0 → u^3 + 2*u^2 + 3*u + 4 > 4 ≠ 0 For real x, no need to solve the cubic.

[@phandinhthanh2295]

No, sin(x) and cos(x) can both be negative on [0; 2*pi).

[@albertmcchan]

To get sin(x)^5 + cos(x)^5 = 1, LHS 2 terms must be non-negative, because both terms at most reached 1. Because 5th power is odd function, this implied sin(x) and cos(x) also non-negative.

[@phandinhthanh2295]

@@albertmcchan Ok, I've got it.

[@veena7274]

The problem can be done in a simpler manner if we consider u = sinx ^2 and the expression becomes x^3 + (1 -x)^3. It will work out to be an easy quadradic on x.

[@SyberMath]

Are you sure? 🤔

[@veena7274]

@@SyberMath Sorry! I am such an idiot

[@mcwulf25]

I factored out (s+c) and reduced what's left to sc(sc-1) using the sum of squares= 1 a couple of times. Both factors must be 1 or both -1. After that it's easy substitutions.

[@mcwulf25]

Using this method it's clear that the only solutions are when s=0,c=1 and s=1,c=0. Two other solutions drop out but these have c or s > 1.

[@davesimms8825]

Holy crap that took a long time to solve something that was obvious just by looking.

[@SyberMath]

Hmm

[@lequangvan6934]

6:00 sin^3x + cos^3x = (3u - u)/2 (Sin^3x +cos^3x)(sin^2x + cos^2x) = (3u - u)/2

[@rajakumar9377]

Put x equals 45° ......then this expression gives value < 1 Hence given expression is false or invalid

[@frankhofmann7263]

I solved this equation by finding the maxima of f(x) = sin^5(x) + cos^5(x) between [0;2pi[ , f'(x) = 0 --> Maximum at x1 = 0 and x2 = pi/2, f(0) = 1 and f(pi/2) = 1. To my mind this is much easier.

[@AtomicMathematics]

This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution kzbin.info/www/bejne/rZutfJuFj56SgZo

[@caoainguyenang1007]

This formal way is so complicated Sin^5

[@marcosgoldin5623]

Muy buen trabajo. Saludos desde Peru

[@SyberMath]

Thank you! Greetings from the United States!

[@emilaz5493]

Math is not difficult but it is needs time to think for resolving every equation!

[@SyberMath]

Math is not hard! You just need to memorize 1578 formulas! 🤣

[@srividhyamoorthy761]

Please explain asum with cubic formula and explain the cubic formula in detail and all the forms of the eqn I am notable to find any video helping me

[@godiswatching8110]

From relations 1=sin(x)^5+cos(x)^5

[@MC_Transport]

5:22 You forgot the negative sign in front of 3u So It will be (-3u-u³)/2 not (3u-u³)/2

[@quanghuy8970]

Thanks ad.

[@tmacchant]

Squaring both side of sin(x)+cos(x)=1 gives extrareneaus solutions as mentioned in the end of part the video. Using relation of sin(x)+cos(x)=√2･sin(x+π/4)=1 does not give extrareneaus solutions.

[@jarikosonen4079]

If it can be "seen" that sin^5(x)+cos^5(x)

[@SyberMath]

Very nice! I'll probably redo this and use your idea 🤩

[@neeruyadav9947]

Put x=0degree then question become easy I think

[@brentsoper345]

So I solved for an expression in terms of tanx. I had to allow the two real solutions of x which are 0 and pi/2. This gave a nifty quadratic where tanx ^5= b and b^2-1b +1 =0 I used the quadratic formula and de Moivres theorem to end up with two solutions 1 cis plus or minus pi/ 15 In the range allowed I found two complex solutions. For the angles pi/15 and 29pi/15

[@brentsoper345]

tanx^5 gave two solutions from quadratic which are (1plus or minus root3)/2 nice little angles of 60 degrees on complex number diagram.

[@Mathokgood]

Phương trình về các hàm số lượng giác. Giải bởi đặt ẩn phu. Cảm ơn.

[@SyberMath]

Không vấn đề gì!

[@RhythmofSumit]

What application is this? I mean any software is there? Can you tell me? And yes nice elaboration

[@SyberMath]

Notability

[@bichvanngoc3150]

Sin ^5x 《 sin ^2 x Sin ^5 x + cos^5 x 《 sin^2 x + cos^2 x =1

[@PATILGameking]

Just took 2 secs to solve the problem. Here take x=90 degrees or pi/2

[@danielalvear3406]

Lo resolví en menos de un minuto recordando q con ciertos ángulos una función es cero y la otra es uno para el mismo valor del ángulo Por lo tanto obtuve como resultado 0, 90 y 360

[@ku6r1ck]

actually, all the point of (cost, sint) lies on unit circile x^2+y^2 = 1, and it implies that all solutions of equation cos^n(x) + sin^n(x) = 1 also must lie on the trivial vertexes, (1,0) and (0, 1).

[@surendraraju5754]

can't we let cos^2(x) = a. This implies sin^2(x) = 1-a. So, we have a^3+(1-a)^3 =1. Solving this gives a=0 or a=1 which implies cos^(2)x=1 or 0. So, x= 0 or pi/2.

[@NurmemetAbliz]

This solution is incorrect . Here is why: x = 0 -> sinx = 0 -> (0)^5 + 1^5 ; x = pi/2 -> cosx = 0 -> 1^5 + 0^5; And 0^5 has no meaning . It should not be permitted in the equation .

[@malabikasaha2452]

Solution using trigonometric identities is much shorter and quite easy-half page.

[@mustafaerenguzel]

Vay be, harikaydı. Teşekkür ederiz. 🙏💐 Bu arada Türk müsünüz? Selamlar, sevgiler :)

[@SyberMath]

Rica ederim. Ben de teşekkür ediyorum. Saolun, varolun! 🥰

[@jcfos6294]

L2 or L3 or M1 or M2.... In LMD in university ?

[@ManjulaMathew-wb3zn]

Once you knew the sum of 3rd powers you could have simply multiplied by the sum of squares to get the sum of fifth powers. It avoids a lot of simplifications to end up with the same equation.

[@SyberMath]

I agree

[@rsri2612]

Sin and cos have a phase difference of 90. Each with max value of 1. So the only solution is multiples of 90. However, it's nice to find a hard arithmetic proof.

[@AtomicMathematics]

This is an interesting problem from Math Olympiad. Here the general solution is x=2n⫪, (4n+1)⫪/2. You can check this video for the solution kzbin.info/www/bejne/rZutfJuFj56SgZo

[@AtomicMathematics]

Thank you sharing this nice problem. There is an tricky method to solve this question. Here the general solution is x=2n⫪, (4n+1)⫪/2.

[@SyberMath]

Np. Thanks

[@peterbanh1364]

x=2nPi or x=(2n+1)Pi where n is any integer.