brainlets get one hour to solve this or else /sci/ gets deleted and you all commit seppuku:
[eqn]\sin \left(\frac{1}{x}\right)=\sin (x)[/eqn]
>>8312518
x=1
>>8312526
cuck
Sinx? Sinks? What about them?
>>8312532
(You)
could you really not solve this homework by yourself
>>8312518
Are you really that retard?
>>8312518
x= infinity
>>8312551
It's kinda hard.
x = ± 1
x = +- sqrt( pi^2 n^2 + 1) - pi n
where n is an integer
x = -z*π ± sqrt( (z*π)^2 + 1)
where z is any integer
>>8312607
ah, beat me to it
We wish to solve [math]\sin \frac1x = \sin x[/math]. Because [math]\sin(x + 2\pi) = \sin x = \sin \bigl( \frac\pi2 - x \bigr)[/math], there are two cases: [math]\frac1x = x + 2 \pi k[/math] and [math]\frac1x = \frac\pi2 - x + 2 \pi k[/math], both of which are for arbitrary [math]k \in \mathbb Z[/math].
In the first case, WLOG [math]x^2 - 2\pi k x - 1 = 0[/math], which we solve to find [math]x = \pi k \pm \sqrt{\pi^2k^2 - 1}[/math]. The second case is similar and more disgusting. So fuck writing that out.
Also, Wolfram|Alpha would have done this homework for you just as well: http://www.wolframalpha.com/input/?i=sin(x)+%3D+sin(1%2Fx)
>>8312569
>difficult
please go away
x = -sqrt(π^2 n^2+1)-π n, n element Z
I don't remember doing this in high school
Sin(1/x) = sin(x)
Sin^-1(sin(1/x)) = sin^-1(sin(x))
1/x =x
1 = x^2
+/-1 = x
>>8312635
>[math]\sin[/math] is invertible
>look see i inverted it: [math]\sin^{-1}[/math]
you might wanna retake precalc there buddy
>>8312652
WOW you dont even know inverse from invert!
I love it, keep going
>>8312652
Sin(x)=y
Arcsin(y)=x
Does your calculator say arcsin or sin^-1 ??
>>8312659
Okay
>Inverse trigonometric functions
>From Wikipedia, the free encyclopedia
> (Redirected from Arcsine)
>
>In mathematics, the inverse trigonometric functions (occasionally called cyclometric functions[1]) are the inverse functions of the trigonometric functions (with suitably restricted domains).
>with suitably restricted domains
Is that enough or should I go on?
>>8312670
Im glad you finally learned something
>>8312678
wow teach thanks, i just have one more question
how do i not lose solutions to the original equation when i restrict the domain of [math]x[/math]?
because this is a lossy solution method, of course
>>8312685
Add n(pi)
>>8312696
2n*pi you fucking pleb
I'm more interested in all possible answers.
sin(1/x) will approach 0 as x goes to infinity, and sin(x) will intersect these values for an infinite amount of values, although they will be slightly further apart each time. The separation of values will approach the finite limit of pi, but what values do they take before the limit?
>>8312621
sin(x)=sin(pi-x) not pi/2 so
1/x=2kpi+x or (1+2k)pi-x
Also your solution should hold for k=0, it should be +1 in the root.
>>8312710
2pi gets you to the same place!
I really like seeing you try, please continue!