I want to find general solutions ($\theta$ is real) to the equation $$\sin(\pi\cos\theta)=\cos(π\sin\theta).$$
What I tried (didn't lead to much):
$$\cos(\pi\sin\theta)=\cos(π/2-\pi\cos\theta).$$
I eventually found $\theta,$ but it does not match the answer given in my textbook, which is $$4\theta=(2m+1)\pi \pm \arccos (1/8),$$ where $m$ is an integer. I always struggle with getting the right solution sets, and have spent two days on this one.
Your result $\cos(\pi\sin\theta)=\cos(\frac{1}{2}π-\pi\cos\theta)$ implies that $\frac{1}{2}π-\pi\cos\theta = 2n\pi \pm \pi\sin\theta$, so $$\cos \theta \pm \sin \theta = \tfrac{1}{2}-2n.$$ The left hand side is $\sqrt{2}\cos(\theta \mp \tfrac{1}{4}\pi)$, so you want to solve $$\cos(\theta \mp \tfrac{1}{4}\pi)=\frac{1}{2\sqrt{2}}.$$ ($n$ must be zero, as all other values of $n$ give values for the cosine outside the range $[-1,1]$.)
Now, if $\cos\alpha = \dfrac{1}{2\sqrt{2}}$, $\cos 2\alpha =-\tfrac{3}{4}$ and $\cos 4 \alpha = \tfrac{1}{8}$, so $$\cos(4\theta \mp \pi)=\tfrac{1}{8},$$ which I think leads to the answer in the textbook.
However, I think the answer in the text book is incorrect. The obvious solution (just find $\theta$) $$\theta =2n\pi \pm \arccos\frac{1}{2\sqrt{2}}\pm \tfrac{1}{4}\pi$$ is the correct one. The problem is that $\cos\alpha = \dfrac{1}{2\sqrt{2}} \implies \cos 4 \alpha = \tfrac{1}{8}$, but the reverse is not true, so the textbook's method introduces spurious roots - in fact, only half of the answers given are actually solutions of the original equation.
Sometimes looking at graph helps.
Looking at the graph, the solutions between $0$ and $2\pi$ are:
$0.4$ , $2$ , $4.3$ and $5.9$. If your values more or less evaluate out to $2\pi \pm$this solution set, your solution set is correct.
The equation
$$\cos(\pi \sin x)=\sin (\pi \cos x)$$
$$\implies \cos(\pi \sin x)=\cos(\pi/2-\pi \cos x)$$
$$\implies \pi \sin x= 2n\pi\pm (\pi/2-\pi \cos x)$$
$$\implies\frac{ \sin x\pm \cos x}{\sqrt{2}}=\frac{2n\pm 1/2}{\sqrt{2}}.$$
Only $n=0$ is valid
$$\implies \cos(x-\pi/4)=\frac{1}{2\sqrt{2}}\tag{1}$$
$$\implies \cos(x+\pi/4)=\frac{1}{2\sqrt{2}}\tag{2}$$
From $(1)$, we get, when $m=0,\pm 1,\pm 2, \pm 3,...$
$$x=2m\pi \pm\frac{\pi}{4}\pm \cos^{-1}\left(\frac{1}{2\sqrt{2}}\right)\tag{3}.$$
Below here is the plot of $f(x)=[\cos(\pi \sin x)-\sin(\pi \cos x)]$ whose period is $2\pi$. Using $(3)$ we get the first four roots of $(*)$ in $(0,2\pi)$ as $x=0.4240, 1.9948, 4.2883, 5.8591$
