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From: William Elliot on 29 Jul 2010 06:12
What are the solutions to the equations

sin^-1 x = 1/sin x,

cos^-1 x = 1/cos x?
From: José Carlos Santos on 29 Jul 2010 07:14
On 29-07-2010 11:12, William Elliot wrote:

> What are the solutions to the equations
>
> sin^-1 x = 1/sin x,

Obviously, it has one and only one solution, namely 0.

> cos^-1 x = 1/cos x?

It has one and only one solution, around 0.739085. I doubt that it is
"nice" number.

Best regards,

Jose Carlos Santos
From: Bart Goddard on 29 Jul 2010 07:33
Jos� Carlos Santos <jcsantos(a)fc.up.pt> wrote in news:8bd5sqFrciU1
@mid.individual.net:

> On 29-07-2010 11:12, William Elliot wrote:
>
>> What are the solutions to the equations
>>
>> sin^-1 x = 1/sin x,
>
> Obviously, it has one and only one solution, namely 0.
>
>> cos^-1 x = 1/cos x?
>
> It has one and only one solution, around 0.739085. I doubt that it is
> "nice" number.

I think you're misreading. The first equation has
lots of solutions, and 0 isn't one of them. The
second equation's solution is closer to 0.45.

Rewrite the first equation as sin(csc(x))-x = 0. While
x is hovering around 0, csc(x) goes to +/- infinity,
so sin(csc(x)) visits all the points between -1 and 1
infinitely often. It'll be equal to x infinitely often,
then.



--
Cheerfully resisting change since 1959.
From: Pubkeybreaker on 29 Jul 2010 07:35
On Jul 29, 7:33 am, Bart Goddard <goddar...(a)netscape.net> wrote:
> José Carlos Santos <jcsan...(a)fc.up.pt> wrote in news:8bd5sqFrciU1
> @mid.individual.net:
>
> > On 29-07-2010 11:12, William Elliot wrote:
>
> >> What are the solutions to the equations
>
> >> sin^-1 x = 1/sin x,
>
> > Obviously, it has one and only one solution, namely 0.
>
> >> cos^-1 x = 1/cos x?
>
> > It has one and only one solution, around 0.739085. I doubt that it is
> > "nice" number.
>
> I think you're misreading.  The first equation has
> lots of solutions, and 0 isn't one of them.  The
> second equation's solution is closer to 0.45.
>
> Rewrite the first equation as sin(csc(x))-x = 0.  While
> x is hovering around 0, csc(x) goes to +/- infinity,
> so sin(csc(x)) visits all the points between -1 and 1
> infinitely often.  It'll be equal to x infinitely often,
> then.

Normally, the range of arcsin(x) is -pi/2 to pi/2.....
From: achille on 29 Jul 2010 08:16
On Jul 29, 7:35 pm, Pubkeybreaker <pubkeybrea...(a)aol.com> wrote:
> On Jul 29, 7:33 am, Bart Goddard <goddar...(a)netscape.net> wrote:
>
>
>
> > José Carlos Santos <jcsan...(a)fc.up.pt> wrote in news:8bd5sqFrciU1
> > @mid.individual.net:
>
> > > On 29-07-2010 11:12, William Elliot wrote:
>
> > >> What are the solutions to the equations
>
> > >> sin^-1 x = 1/sin x,
>
> > > Obviously, it has one and only one solution, namely 0.
>
> > >> cos^-1 x = 1/cos x?
>
> > > It has one and only one solution, around 0.739085. I doubt that it is
> > > "nice" number.
>
> > I think you're misreading.  The first equation has
> > lots of solutions, and 0 isn't one of them.  The
> > second equation's solution is closer to 0.45.
>
> > Rewrite the first equation as sin(csc(x))-x = 0.  While
> > x is hovering around 0, csc(x) goes to +/- infinity,
> > so sin(csc(x)) visits all the points between -1 and 1
> > infinitely often.  It'll be equal to x infinitely often,
> > then.
>
> Normally, the range of arcsin(x) is -pi/2 to pi/2.....

arcsin( 0.9440390666116 ) ~ 1.234668961508658 ~ 1/
sin(0.9440390666116) ???
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