Finding the nontrivial zeros of Tan x = x

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I can do the first part no problem.

I then drew the graph, am I right in saying there is an infinite sequence because the lines intersect an infinite amount of times, because tan is periodic and has vertical asymptotes?

I have no idea about showing why the first non rivial zero is bounded like that. I would have thought the trivial zero was at the origin, and the first non trivial was in the range of 0<lambda<pi/2.

What am I not understanding here?

Thanks

edit:

oh I think I'm suppose to be looking for solutions in the range of pi and 3pi/2?
 
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Firepanda said:
oh I think I'm suppose to be looking for solutions in the range of pi and 3pi/2?

...because \lambda is a positive number, so the first non-trivial zero of the equation will be in the next quadrant where \tan \lambda is positive, and you're after \lambda^{2}.

[EDIT: On thinking about this a little more, we could also go in the negative direction, since \tan \lambda and \lambda both have odd symmetry and we're looking for solutions for \lambda^{2}... But it is easier to think about going in the positive direction.]

BTW, I think the problem-poser means the first non-trivial zero of the function (\tan \lambda ) - \lambda ; that last statement reads a little strangely...
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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