how do you calculate the inersection points of sqrt(sin(x)) and 2x/pi

Question

anonymous · Answer

please

psymon · Answer

I have the answers, just not sure how youd get them by hand to be honest. Im trying to think of something.

anonymous · Answer

yeah same :L but thanks for the help

anonymous · Answer

i legit have no idea how wolfram does it (only if they showed the working)

anonymous · Answer

is there a way to work back? from (0,0) and (1,pi)

psymon · Answer

Well, the only thing I could think of was this. And it may not be a legitimate way, its just a logic way. 

Let u = 2x/pi. If I solve for x, I get x = (pi/2)u

$$\sqrt{\sin (\frac{ \pi }{ 2 }u)}= u$$
Not that this gives you much, but it makes picking values easier to do with this simplified.

anonymous · Answer

yeah true. 
there must be some advanced concept i have no idea about.

but yeah thanks heaps for your help!

dumbcow · Answer

i would use Newtons method here http://en.wikipedia.org/wiki/Newtons_method $$f(x) = \sqrt{\sin(x)} - \frac{2}{\pi}x$$ $$f'(x) = \frac{\cos(x)}{2\sqrt{\sin(x)}} -\frac{2}{\pi}$$ $$x_{n+1} = x_n -\frac{f(x)}{f'(x)}$$

psymon · Answer

Ah, so you pretty much would be doing an approximation. I guess I assumed there was some definite way, but newtons method if you dont know what else to do o.o

dumbcow · Answer

oh i am assuming there is no algebraic way to solve this, at least no way i have ever seen

psymon · Answer

Yeah, I dont think there is either, I guess I was just assuming maybe some fancy substitution, lol. If the answers weren't as clean as they are then yeah, would have to just assume newtons.

anonymous · Answer

Umm, we know $x=0$ just by looking.

anonymous · Answer

Then there is $\pi/2$

anonymous · Answer

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And those are the only points where it intersects.