How to solve system of three trigonometric equations:
$(\sin x)^2 (\cos y)^2 = 4 \cos x \sin y\tag1$
$(\sin y)^2 (\cos z)^2 = 4 \cos y \sin z \tag2$
$1- \sqrt{\sin z}(1+\sqrt{\cos x})=\sqrt{\frac{1-\sin y}{1+\sin y}}\tag3$
and to verify that
$\sin x=\sqrt{2}(\sqrt{2}+1)\sqrt{\sqrt{10}-3}(\sqrt{5}-2)\\ \sin y=(\sqrt{2}-1)^2(\sqrt{10}-3)\\ \sin z=(2\sqrt{2}+\sqrt{5}-\sqrt{12+4\sqrt{10}})^2(\sqrt{5}+\sqrt{2}-\sqrt{6+2\sqrt{10}})^2$
not is the only solution, but that many others exist?
The simplest solution is:
$\sin x=2^{5/4}\big(\sqrt{2}-1\big)$
$\sin y=\frac{1}{\sqrt{2}}$
$\sin z=3-2\sqrt{2}$.
Three other solutions are:
1)
$\sin x=\sqrt{2\sqrt{2}-2}$
$\sin y=\sqrt{2}-1$
$\sin z=5+4\sqrt{2}-\sqrt{56+40\sqrt{2}}$,
2)
$\sin x=\frac{\sqrt{2}}{2}$
$\sin y=3-2\sqrt{2}$
$\sin z=\big(\sqrt{2}+1\big)^{2}\big(2^{1/4}-1\big)^{4}$
3)
$\sin x=\frac{2^{5/4}a^{2}c. d^{6}\varphi^{3}}{b^{2}}$
$\sin y=\frac{d^4}{\sqrt{2}\varphi^{4}}$
$\sin z=a^{4}b^{4}c^{2}\varphi^{6}$
where
$$a=\sqrt{2}-1$$ $$b=5^{1/4}-\sqrt{2}$$ $$c=\sqrt{10}-3$$ $$d=\frac{5^{1/4}-1}{\sqrt{2}}$$
$\varphi$ is the golden ratio.