I Solved A Homemade Trigonometric Equation 😊

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Опубликовано:

9 янв 2024

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Комментарии : 30

@SyberMath 6 месяцев назад

Sorry for the confusion. I fixed the thumbnail. It now reflects the actual problem: sinx - icosx = 1 I hope you enjoyed it. I'm aware that 1^x = e does not have a solution. I will fix it when I get a chance. Here's the video in case you want to see the problematic solution: 😮😂 ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lsHnl-nyAh4.html

@yoav613 6 месяцев назад

The answer fot the thumbnail and for the 10 first seconds is x=pi/2-i+2pin😊

@SyberMath 6 месяцев назад

What is the pin number? 😁😂

@yoav613 6 месяцев назад

@@SyberMath ask wolfarm😃

@grink_coolhoznik 6 месяцев назад

sinx-i*cosx=e. Producting both sides with i: i*sinx-(i^2)cosx=ei; cosx+i*sinx=ei. But the left side also equals e^(ix), so ei=e^(ix). i=e^i(π/2+2πk) (k is an integer), so ei=e*e^i(π/2+2πk)=e^(1+i(π/2+2πk)). e^(1+i(π/2+2πk))=e^(ix); e^((π/2+2πk)-i)=e^x; x=π/2+2πk-i

@seanfraser3125 6 месяцев назад

sinx-icosx = -ie^(ix). So -ie^ix = e. Thus e^ix = ie = e^(1+ipi/2) From here ix= 1+i(pi/2 + 2npi), where n is any integer, so x= pi/2 + 2npi - i.

@ianfowler9340 6 месяцев назад

BTW you can still do e. x = [ ln(i) + 1 ]/i

@yoav613 6 месяцев назад

Sinx-icosx=e^i(x-pi/2) and frome here it is easy..

@anthonycheng1765 6 месяцев назад

or e^i(3*pi/2+x),......

@calculus988 6 месяцев назад

I wish i had Sybermath's mind. I could easily understand everything 😅 Trigonometry is a hard subject. Everyone, wish me the best in my personal study in Trigonometry.

@SyberMath 6 месяцев назад

❤❤ Thank you for the kind words! I wish you best of everything. Do not worry. You'll get there

@ianfowler9340 6 месяцев назад

Another path: (as also seen below) sin(x) - icos(x) = 1 (multiply by i) cos(x) + isin(x) = i e^(ix) = i ix = ln(i) x = ln(i) / i

@SyberMath 6 месяцев назад

A beautiful way to solve it. 😍😍 How did that escape me? 😮😂

@ianfowler9340 6 месяцев назад

@@SyberMath Don't beat yourself up too much. It happens to us all. Cheers.

@scottleung9587 6 месяцев назад

Nice!

@seegeeaye 6 месяцев назад

From the given equation sinx-icosx=1, solve the simultaneous equations sinx=1 and cosx=0, we have x=pai/2 +2npai

@shreeniwaz 4 месяца назад

Thank God now we have the technology to solve such problems..😅

@kassuskassus6263 6 месяцев назад

What would happen if we multiply both sides by i ? cos x + isin x = i, with i = exp (i(pi/2 + 2kpi)), then x = pi/2 + 2kpi.

@SyberMath 6 месяцев назад

A beautiful way to solve it. 😍😍

@lucaswarnke3668 6 месяцев назад

e=e^(-ii)=cos -i+isin -i

@xsimox13 6 месяцев назад

I don’t understand why sinx - i cosx = 1 if you say at first sinx - i cosx = e that’s not the same is it ?

@SyberMath 6 месяцев назад

I messed up and now changed the thumbnail

@tonyhague5357 6 месяцев назад

Goodness. Just equate real and imaginary parts; sin(x)=1, cos(x)=0. x=pi/2 + n×2pi. Done.

@SyberMath 6 месяцев назад

Good thinking!

@klementhajrullaj1222 6 месяцев назад

And why not theta=pi/2+x?!

@ManjulaMathew-wb3zn 4 месяца назад

Actually I’m surprised you overlooked the obvious as you generally have a great pair of eyes. sinx-I cosx=1=1+i0 Comparing real parts sinx=1 x=2PIn+PI/2 You can get the same result by comparing imaginary parts. cosx=0 Interestingly if you change the - sign to + the answer would be the same. 😊

@SyberMath 4 месяца назад

nice

@dariosilva85 6 месяцев назад

Your solution is wrong. x=pi/2 + 2n*pi is not the correct solution of sinx - icosx = e. Put in pi/2 into the original equation, you get that 1=e. The correct solution is x = pi/2 - (2n*pi +1)i. You went wrong at the line when you said e^ix = i. It should be e^ix = ie.

@August-ol5uh 6 месяцев назад

His solution is correct, he was solving the equation: sinx - icosx = 1 not sinx - icosx = e , he made a mistake in writing the question but he corrected it in the video and in the comment section