An Exponentially Trigonometric Equation

[@SyberMath]

Here's the graph I forgot to show you!!! 😜😮😁 ibb.co/zQc9sRV

[@scottleung9587]

Cool!

[@TypoKnig]

For real x, sinx and cosx are between -1 and 1, so cannot be an even integer. Finding the complex values of cox^2(x) = 2n and of sinx=2n would be interesting!

[@coshy2748]

A correction to your approximation at 7:06 min 1 - 1.6.... ~ - 0.6 (not - 0.4)

[@CarolosNiesha]

Thanks for the breakdown! I have a quick question: My OKX wallet holds some USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). How should I go about transferring them to Binance?

[@gregevgeni1864]

Merry Christmas!

[@SyberMath]

Merry Christmas! Happy Holidays!!! 🎁🎄 😍

[@BOBPERIO2]

Merry Christmas and Happy New Year to @SyberMath and @aplusbi! Thank you for all your great work over the past year. I look forward to your posts each and every day. You’re the best!

[@SyberMath]

Merry Christmas and Happy New Year to you too! 🎄 Thank you for the kind words! 😍

[@sadnankhan3042]

At 5 :30, equation s^2+s-1=0

[@zawatsky]

Сводится оно до более интуитивно понятной системы: (b/a)^(a²/c²)=(a/b)^(b/c), при a²+b²=c², вот только переменных становится гораздо больше... Сразу видим, что все стороны положительные, но это и так понятно. А степени будут действительны - это тоже ясно. Первый случай: b/a=a/b=1, либо b/a=-a/b=±1 и степень чётная, либо b/a=a/b=-1 и степень нечётная. Подходит только первый вариант - равнобедренный прямоугольный треугольник, т. е. x=45°. Второй случай: (a/b)^(a²/c²)=(a/b)^(-b/c)⇔a²/c²=-b/c. Вернём к первоначальному виду: cos²x=-sinx=1-sin²x⇔sin²x-sinx-1=0. Заменим на t: t²-t-1=0, D=5, синус равен золотому сечению.

[@bbajr]

i know here tanx = 0 is not considered as it fails both cases. but should we also consider that for general cases?

[@Blaqjaqshellaq]

If x=0, we have a removable discontinuity. Redefine the equation as tan(x)^[cos(x)^2+sin(x)]=1 and 0 will also be a solution.

[@vaggelissmyrniotis2194]

X=arcsin((5^(1/2)-1)/2) and x=kπ+π/4, κ is an integer. Merry christmas!!

[@SyberMath]

Merry Christmas and Happy New Year to you too! 🎄

[@shaileshs.3177]

By observation X = π/4 is a solution.

[@dan-florinchereches4892]

I am going to do the Weierstrass substitution I guess t=TAN(X/2) tanx= 2t/(1-t^2) Sinx=2t/(1+t^2) cosx=(1-t^2)/(1+t^2) So (2t/(1-t^2))^(1-t^2)^2/(1+t^2)^2=((1-t^2)/2t)^(2t/(1+t^2)) => (2t/(1-t^2))^((1-t^2)^2/(1+t^2)^2+2t/(1+t^2))=1 So we can have the base =1 or exponent =0 Case 1: 2t/(1-t^2)=1 t^2+2t-1=0 t1,2=(-2+-√8)/2=-1+-√2 Case 2 (1-t^2)^2/(1+t^2)^2+2t/(t^2+1)=0 t^4-2t^2+1+2t(t^2+1)=0 t^4+2t^3-2t^2+2t+1=0 divide by t^2 t^2+1/t^2+2(t+1/t)-2=0 (t+1/t)^2+2(t+1/t)-4=0 t+1/t = (-2+-√20)/2=-1+-√5 Lots of solutions looks like then we do arctangent and double the values

[@giuseppemalaguti435]

(cosx)^2lntgx=-sinxlntgx...1soluzione tgx=1...2sol 1-(sinx)^2=-sinx...sinx=(1+√5)/2...sinx=(1-√5)/2..non credo siano accettabili queste ultime

[@Caio-Ann]

this one was easy :P