A Non-Standard Trig Equation | sin(x)=2

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[MrLidless]

My maths teacher gave me this question, personally, but publicly to the rest of the class when I was 12. Next day I replied, correctly, but only got 9/10 as I didn’t include the periodicy of multiple solutions. She was a hard, and smelly, taskmaster.

[luygvkz]

Try this 4^(x)+4x=260

Solve only using using Lambert W function

[scottleung]

Cool - I tried implementing the same method from when u did the vid for cos(x)=2, but for sin the logic is trickier.

[jerzykowalczuk]

That is great. But now since we have the solution for the X, so the sin(x) =2 could you please provide the proof by entering the resolved x into the sin function and see that it gives you the # 2 as a result.
I have made an attempt to do so, but failed miserably :).

[kicorse]

Very nice problem and solution.

I did it by substuting 2 for sin(x) and sqrt(1-2^2) for cos(x) in the Euler formula, but I wasn't confident that this was a valid approach when x is complex. I got the same answer as you though, so I'm guessing it is?

[robfrohwein]

Remarkable, I'm going to try to repeat this 😃

[lawrencejelsma]

An analysis of the answer in Electrical Engineering that could use the results but only for "black box theories!" The Z part = π/2 (+ maybe 2nπ redundant important) - ln(2 + √3)i = real + i(imaginary real). Is further advanced by e^(iZ) as (i)[e^(2 +/-√3)] by noticing all π/2 + 2nπ redundancies in explaining complex number "i" . In simpler terms unless a domain requests it π/2 is the domain angle for "i" to eliminate range redundant iSin(∆) information in e^(iZ). What we see for that Z is a mapping of "i" of magnitude|1| transform by a complex number angle to "i" times +/-(2 + √3) to mean: attenuation of e^(-(ln2 + √3)) of the imaginary "i" if negative. or gain of e^(ln2 + √3) of the magnitude in "i"

[bhavya]

algabreacially this is pretty straightforward, but what does the "i" in an angle signify geometrically?

[Alexey_Alex]

В военное время значение синуса может достигать четырёх!

[sdspivey]

Is there an integer that sin(x) can equal, that also has an integer complex solution? Obviously ignoring 1, 0, and -1.

[actualRocketScientist]

My question is how much coffee do you drink?

[wbevqsx]

В твоєму розв'язку є протиріччя: Х в результаті отримується як комплексне число, а насправді Х як аргумент комплексного числа ні в яких випадках не може бути комплексним, а лише дійснимчислом!!! . Тому свій псевдорозв'язок можеш засунути собі кудись поглибше і не мороч людям

[rob]

sin x = 1/(2i)(exp(ix) - exp(-ix)) = 2
let exp(ix) = u
u - 1/u = 4i
u^2 - 4iu - 1 = 0
(u - 2i)^2 + 3 = 0
u - 2i = +/- i.sqrt(3)
u = 2i +/- i.sqrt(3)
ix = ln(2i +/- i.sqrt(3))
= ln(i(2 +/- sqrt(3)) = ln i + ln(2 +/- sqrt(3))
ln i = i(pi/2 + 2ni.pi) = i.pi/2(1 + 4n)
x = pi/2(1 + 4n) - i.ln(2 +/- sqrt(3))

[DictoDictov]

«В военное время значение синуса может достигать двух»

[eckhardfriauf]

Awesome. Eck-stremely ix-ish. But where is the magenta? Did I miss it? 😕😉

[seanfraser]

Obviously there are no real solutions. For convenience let me rewrite using z:
sinz = 2

We could directly use the formula for sinz and that would be fine, but I find it easier to break it down by real and imaginary parts. If z = x + iy where x and y are real, then we have
sinx*coshy + icosx*sinhy = 2

This gives us a system of equations

sinx*coshy = 2
cosx*sinhy = 0

Starting with the second equation, we either have sinhy = 0 or cosx = 0. If sinhy = 0, then y = 0, so coshy = 1. Thus the first equation becomes sinx = 2, which has no solutions.

Thus cosx = 0, or x = pi/2 + n*pi. Plugging this into the first equation gives us (-1)^n*coshy = 2. Since coshy > 0 for real y, we see that n must be even. Now all we need to do is solve coshy = 2.

We can use the known formula for inverse cosh (which is derived from the quadratic formula - I'll let the comment reader work it out) to get y = ln(2+sqrt(3)). Of course, we also need to include the negative solution since cosh is even, so y = +- ln(2+sqrt(3) = ln(2+-sqrt(3))

Thus our final set of solutions is

z = (4n+1)*pi/2 +- i*ln(2+sqrt(3))

where n is any integer.