Find the smallest positive number $p$ for which the equation $\cos(p\sin x)=\sin(p \cos x)$ has a solution $x\in[0,2\pi].$

Question

Find the smallest positive number $p$ for which the equation $\cos(p\sin{x})=\sin(p\cos{x})$ has a solution $x$ belonging $[0,2\pi]$. I am not able to solve this problem. Please help me.

Answer 1

It must be that $p\sin{x}+p\cos{x}=\dfrac{\pi}{2}$ with, $p=\dfrac{\pi}{2(\sin{x}+\cos{x})}$.

So, to minimize $p$, $\sin{x}+\cos{x}$ must be maximized.

$\sin{x}+\cos{x}=\sqrt{2} \sin\left(x+\dfrac{\pi}{4}\right)$, which is maximized when $\sin\left(x+\dfrac{\pi}{4}\right)=1$ at $x=\dfrac{\pi}{4}, \dfrac{7\pi}{4}$.

Hence, $p=\dfrac{\pi}{2\sin\left(\dfrac{\pi}{2}\right)}$.

$p=\dfrac{\pi}{2\sqrt2}$.

Hence, $x=\dfrac{\pi}{4}, \dfrac{7\pi}{4}$ in the interval $[0, 2\pi]$.

Answer 2

Given two angles $\alpha$ and $\beta$ one has $$\cos\alpha=\sin\beta=\cos(\beta-{\pi\over2})$$ iff either $$\alpha=\beta-{\pi\over2}+2k\pi\tag{1}$$ or $$\alpha=-(\beta-{\pi\over2})+2k\pi,\quad{\rm i.e.}\quad \alpha+\beta={\pi\over2}+2k\pi\ .\tag{2}$$ In our case $\alpha=p\sin x$ and $\beta=p\cos x$, so that $(1)$ is equivalent with $$p(\cos x-\sin x)={\pi\over2}-2k\pi\ ,$$ so that we have to make sure that the equation $$p{2\over\sqrt{2}}\sin({\pi\over4}-x)={\pi\over2}-2k\pi$$ has a real solution $x$. This is the case if $${2\over\sqrt{2}}p \geq|{\pi\over2}-2k\pi|$$ for a suitable $k$, and the smallest positive $p$ that satisfies this is $p={\pi\over 2\sqrt{2}}$. The resulting solution $x$ of the original equation is then $x={7\pi\over4}$.

The case $(2)$ is similar and has the same outcome for $p$ (so that this is the definitive solution of the problem); the corresponding $x$ is then ${\pi\over4}$.

Answer 3

I was just made aware of this question in chat, so I thought I'd post the work I did there in case it's useful.

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The equation $$ \cos(p\sin(x))=\sin(p\cos(x))\tag1 $$ is equivalent to $$ \sin\left(\frac\pi2-p\sin(x)\right)=\sin(p\cos(x))\tag2 $$ which, since $\sin(x)=\sin(\pi-x)$, has two solutions: $$ \frac\pi2-p\sin(x)=p\cos(x)\tag3 $$ and $$ \frac\pi2+p\sin(x)=p\cos(x)\tag4 $$ $(3)$ is equivalent to $$ \frac\pi{2p}=\sqrt2\cos\left(x-\frac\pi4\right)\tag5 $$ and $(4)$ is equivalent to $$ \frac\pi{2p}=\sqrt2\cos\left(x+\frac\pi4\right)\tag6 $$ If $\left|\frac\pi{2p}\right|\gt\sqrt2$, there can be no solutions since $|\cos(x)|\le1$.

If $\left|\frac\pi{2p}\right|\le\sqrt2$, then the the solutions to $(1)$ are $$ x\equiv\pm\cos^{-1}\left(\frac\pi{2\sqrt2\,p}\right)\pm\frac\pi4\pmod{2\pi}\tag7 $$

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Thus, the smallest positive $p$ is $\frac\pi{2\sqrt2}$, for which, $(1)$ has the solutions $x\equiv\pm\frac\pi4\pmod{2\pi}$.