Question

The smallest positive integral value of $p$ for which the equation $cos\left(psinx\right)=sin\left(pcosx\right)$ has a solution in interval $[0,2\pi ]$

Also find possible number of solutions in the given interval

Answer 1

$\cos \left(p\sin x\right)=\sin \left(p\cos x\right)$

$\Rightarrow \cos \left(p\sin x\right)=\cos (\frac{\pi }{2}-p\cos x)$

$\Rightarrow p\sin x=\frac{\pi }{2}-p\cos x$

$\Rightarrow p(\sin x+\cos x)=\frac{\pi }{2}$

$\Rightarrow \sin x+\cos x=\frac{\pi }{2p}$

$\Rightarrow |\sin x+\cos x|\le \sqrt{2}$

$\therefore \frac{\pi }{2p}\le \sqrt{2}$

Or $p=\frac{\pi }{2\sqrt{2}}$

$\Rightarrow \sin 5\theta \cos 3\theta =\sin 9\theta \cos 7\theta$

$\Rightarrow 2\sin 5\theta \cos 3\theta =2\sin 9\theta \cos 7\theta$

$\Rightarrow \sin 8\theta \sin 2\theta =\sin 16\theta \sin 2\theta$

$\Rightarrow \sin 16\theta =\sin 8\theta$

$\Rightarrow 16\theta =n\pi \pm {(-1)}^{n}8\theta$

$\Rightarrow 16\theta =2m\pi +8\theta$

If $n$ is even and $n=2m$

$\Rightarrow 8\theta =2m\pi$

$\Rightarrow \theta =\frac{m\pi }{4}$

$\&16\theta =(2m+1)\pi -8\theta$

If $n$ is odd and $n=2m+1$

$\Rightarrow 24\theta =(2m+1)\pi$

$\Rightarrow \theta =\frac{(2m+1)}{24}\pi$

Thus, $\theta =\frac{m\pi }{4},\frac{(2m+1)}{24}\pi ,$ where $m\in z$

$=0,\frac{\pi }{4},\frac{\pi }{2},\frac{\pi }{24},\frac{7\pi }{24},\frac{3\pi }{24},\frac{11\pi }{24}$

$\therefore$ Total no. of solutions $=9.$

Hence, the answer is there are nine solutions.