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Find the smallest number p for which the equation $\cos (p\sin (x))=\sin(p\cos (x))$ has a solution.

x belongs to $[0,2\pi] $

Any hints for this please.Don't know how to proceed.

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  • $\begingroup$ Do you mean the smallest $p>0$ for which the equation has a solution in $x$? $\endgroup$ – almagest May 29 '16 at 13:46
  • $\begingroup$ @ChristianBlatter sorry.Corrected it $\endgroup$ – user220382 May 29 '16 at 13:49
  • $\begingroup$ Isn't the answer practically already in the answer of your previous question (math.stackexchange.com/questions/1803639/…)? $\endgroup$ – Jack D'Aurizio May 29 '16 at 14:24
  • $\begingroup$ Eh :-P....you noticed that before I did :-)..thanks BTW...got it now $\endgroup$ – user220382 May 29 '16 at 16:42
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We have $$\cos(p\sin x)=\sin (p\cos x)$$ $$\implies \cos (p\sin x)=\cos(\pi/2-p\cos x)$$ $$\implies p\sin x=2k\pi+\pi/2-p\cos x$$ Where $k \in \mathbb Z$ $$\implies p(\sin x+\cos x)=(4k+1)\pi/2$$ For finding smallest $p$, $k=0$, $$\implies p(\sin x+\cos x)=\pi/2$$ Now, $\sin x+\cos x\leq \sqrt {2}$, $$\implies p \geq \frac{\pi}{2\sqrt2}$$

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    $\begingroup$ I think there is a mistake while transiting from second to third step.The solution should be given by $2k\pi+ $ or $2k\pi-$ .You didn't take the negative sign. $\endgroup$ – user220382 May 29 '16 at 13:55
  • $\begingroup$ When $p=\frac\pi{2\sqrt2}$, the equation is solved by $x=\pm\frac\pi4+2k\pi$. $\endgroup$ – Akiva Weinberger May 29 '16 at 13:55
  • $\begingroup$ @SanchayanDutta $k$ can be negative as well, so the $2k\pi-$case happens when $k<0$ $\endgroup$ – Nikunj May 29 '16 at 13:57
  • $\begingroup$ If $x$ solves the equation, so does $-x$. So we can, without loss of generality, replace $x$ by $-x$ (and $k$ by $-k$) in that case to turn it into the $2k\pi+$ version. @SanchayanDutta (Graphing confirms that $\frac\pi{2\sqrt2}$ is indeed the answer.) $\endgroup$ – Akiva Weinberger May 29 '16 at 13:59
  • $\begingroup$ I have a doubt.Why did you take k=0, k could be negative as well right? So lower bounds should be possible for p... $\endgroup$ – user220382 May 29 '16 at 14:02

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