you have $$\sin x \cosh y = 3, \cos x \sinh y = 0 $$ take the second equation. you have $$\sinh y = 0 \to y = 0 \\ \cos x = 0, x = \pm \pi/2 +2k\pi $$
putting $y = 0,$ in the first equation gives $\sin x = 3$ which has no real solution. we are now left with $$x = \pm \pi/2 +2k\pi \to \cosh y = \pm 3.$$ since $\cosh y \ge 1,$ we only need to solve $$\cosh y = 3 \to e^{2y} - 6e^y+1=0\to e^y = 3\pm2\sqrt2 \to y = \ln(3\pm2\sqrt2)$$
the solutions are $$\sin^{-1}(3)=\pm \pi/2 +2k\pi+i \ln(3\pm2\sqrt2).$$
I'm going to have a look on your try, and then add a claim and an example which might help you.
From
$$-(\cos^2x+\sin^2x)\leqslant 2\cos x\sin x\leqslant \cos^2x+\sin^2x\tag1$$
and
$$\small -(\cos^2 x+\sin^2 x+\sin^2\alpha)\leqslant 2\cos x\sqrt{\sin^2x+\sin^2\alpha}\leqslant \cos^2x+\sin^2x+\sin^2\alpha\tag2$$
you got
$$-2-\sin^2\alpha\leqslant 2f(x)\leqslant 2+\sin^2\alpha$$
and
$$-1-\frac{\sin^2\alpha}{2}\leqslant f(x)\leqslant 1+\frac{\sin^2\alpha}{2}\tag3$$
To see if $(3)$ is the range of $f(x)$, let us see if there is $x$ such that, for any given $\alpha$, $f(x)=1+\dfrac{\sin^2\alpha}{2}$, i.e.
$$2f(x)=2+\sin^2\alpha\tag4$$
holds.
From $(1)(2)$, $x$ satisfying $(4)$ has to satisfy both
$$2\cos x\sin x= \cos^2x+\sin^2x\tag5$$
and
$$2\cos x\sqrt{\sin^2x+\sin^2\alpha}=\cos^2x+\sin^2x+\sin^2\alpha\tag6$$
$(5)$ is equivalent to
$$(\cos x-\sin x)^2=0\iff \cos x=\sin x\iff x=\frac{\pi}{4}+m\pi\ (m\in\mathbb Z)$$
$(6)$ is equivalent to
$$\cos x\geqslant 0\quad\text{and}\quad \cos^2x=\sin^2x+\sin^2\alpha\iff \cos x\geqslant 0\quad\text{and}\quad \cos(2x)=\sin^2\alpha$$
So, we see that
$$(5)\ \ \text{and}\ \ (6)\iff x=\dfrac{\pi}{4}+2m\pi\quad\text{and}\quad \alpha=n\pi\quad (m,n\in\mathbb Z)$$
This means that there is no $x$ such that, for any given $\alpha$, $f(x)=1+\dfrac{\sin^2\alpha}{2}$ holds.
So, $(3)$ is not the range of $f(x)$.
In the following, I'm going to add a claim and an example which might help you.
In the following, I consider only continuous functions.
Claim : If
$f(x)\leqslant F(x)\leqslant g(x)\quad$ ($f(x),g(x)$ are not necessarily constant functions)
$h(x)\leqslant G(x)\leqslant i(x)\quad$ ($h(x),i(x)$ are not necessarily constant functions)
both $f(x)+h(x)$ and $g(x)+i(x)$ are constant functions
there is $\alpha$ such that $F(\alpha)=f(\alpha)$ and $G(\alpha)=h(\alpha)$
there is $\beta$ such that $F(\beta)=g(\beta)$ and $G(\beta)=i(\beta)$
then, $\operatorname{range}(F(x)+G(x))=\left[f(x)+h(x),g(x)+i(x)\right]$.
Proof :
For $c$ such that $f(x)+h(x)\leqslant c\leqslant g(x)+i(x)$, letting $H(x)=F(x)+G(x)-c$, since we have
$$H(\alpha)=F(\alpha)+G(\alpha)-c=f(\alpha)+h(\alpha)-c\leqslant 0$$
$$H(\beta)=F(\beta)+G(\beta)-c=g(\beta)+i(\beta)-c\geqslant 0$$
we can say that, by the intermediate value theorem, there is $\gamma$ such that
$$H(\gamma)=0\qquad \text{and}\qquad \min(\alpha,\beta)\leqslant \gamma\leqslant \max(\alpha,\beta).\quad\blacksquare$$
Exmaple for which the claim above works :
Find the range of $\cos x+\dfrac{x^2+8\pi x-2\pi^2}{3x^2+6\pi^2}$.
Let $f(x)=\cos x$ and $g(x)=\dfrac{x^2+8\pi x-2\pi^2}{3x^2+6\pi^2}$. Then, $g'(x)=\dfrac{-8 \pi (x-2\pi)(x+\pi)}{3 (x^2 + 2 \pi^2)^2}$ and $\displaystyle\lim_{x\to\pm\infty}g(x)=\frac 13$.
Since $-1\leqslant f(x)\leqslant 1$ and $-1=g(-\pi)\leqslant g(x)\leqslant g(2\pi)=1$, adding these gives
$$-2\leqslant f(x)+g(x)\leqslant 2\tag7$$
Since $f(-\pi)=g(-\pi)=-1$ and $f(2\pi)=g(2\pi)=1$, we can say, by the claim above, that $(7)$ is the range.
Best Answer
Take the $\sin$ of both sides: $$|\cos x|=\sin(\cos^{-1}(|\sin x|))$$ Drawing a triangle for $\theta=\cos^{-1}(|\sin x|)$, you will get that adjacent is $|\sin x|$ and hypotenuse is $1$, so opposite is $\sqrt{1-(|\sin x|)^2}=\sqrt{1-\sin^2x}$. So the RHS becomes: $$\sin(\cos^{-1}(|\sin x|))=\sin\theta=\sqrt{1-\sin^2x}/1=\sqrt{1-\sin^2x}$$ But due to the famous trigonometric identity $$\sin^2x+\cos^2x=1$$ You will find that $\sqrt{1-\sin^2x}=|\cos x|$. This is the exact same thing as the LHS!!!
This indicates that there are infinitely many real solutions in the interval $[0,4\pi]$. In fact, for any $x$ you plug in, both sides of the equation will always be equal to each other!
You can even verify this by graphing $y=\sin^{-1}(|\cos x|)$ and $y=\cos^{-1}(|\sin x|)$ on the same graph in your favorite graphing calculator, and see that they intersect exactly on top of each other!