Question

Find the smallest positive number p for which the equation $cos\left(psinx\right)=sin\left(pcosx\right)$ has a solution $x\epsilon [0,2\pi ]$

• A. $\sqrt{2}\pi /4$
• B. $\sqrt{2}\pi /2$
• C. $\sqrt{2}\pi$
• D. $\pi /6$

Answer 1

The correct option is A $\sqrt{2}\pi /4$

$\cos \left(p\sin x\right)=\sin \left(p\cos x\right)$
$\Rightarrow \cos \left(p\sin x\right)=\cos (\frac{\pi }{2}-p\cos x)$
$\Rightarrow p\sin x=2n\pi \pm (\frac{\pi }{2}-p\cos x)$
$\Rightarrow p\sin x+p\cos x=2n\pi +\frac{\pi }{2}$ or $p\sin x-p\cos x=2n\pi -\frac{\pi }{2}$
$\Rightarrow p\sqrt{2}\left(\sin (x+\frac{\pi }{4})\right)=\frac{(4n+1)\pi }{2}$
Or $p\sqrt{2}\left(\sin (x-\frac{\pi }{4})\right)=\frac{(4n-1)\pi }{2}$
As $-1\le \sin (x+\frac{\pi }{4})\le 1$
$\Rightarrow -p\sqrt{2}\le p\sqrt{2}\sin (x+\frac{\pi }{4})\le p\sqrt{2}$
$\Rightarrow -p\sqrt{2}\le \frac{(4n+1)\pi }{2}\le p\sqrt{2}$ ...(1)
And as $-1\le \sin (x-\frac{\pi }{4})\le 1$
$\Rightarrow -p\sqrt{2}\le p\sqrt{2}\sin (x-\frac{\pi }{4})\le p\sqrt{2}$
$\Rightarrow -p\sqrt{2}\le \frac{(4n-1)\pi }{2}\le p\sqrt{2}$ ...(2)
(2) is always a subset of of first, therefore we have to consider only first
It is sufficient to consider $n\ge 0$ , because for $n>0$, the solution will be same for $n\ge 0$
If $n\ge 0$ $-\sqrt{2}p\le \frac{(4n+1)\pi }{2}\Rightarrow \frac{(4n+1)\pi }{2}\le \sqrt{2}p$
$\sqrt{2}p\ge \frac{\pi }{2}\Rightarrow p\ge \frac{\pi }{2\sqrt{2}}=\frac{\pi \sqrt{2}}{4}$