How to solve a system of 3 trigonometric equations 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.
 A: Note that, if any of $\sin x$, $\sin y$, $\sin z$ are zero, then they all are, so we have the solution
$$\sin x = \sin y = \sin z = 0 \qquad \cos x = 1 \tag{0}$$
(where the latter condition allows $\sqrt{\cos x}$ to be real). On the other hand, if any of $\cos x$, $\cos y$, $\cos z$ are zero, then they all are, and we have
$$\cos x = \cos y = \cos z = 0 \qquad \sin y = \sin z = 1 \tag{00}$$
So, we may assume none of these sines and cosines are zero. Moreover, since $\cos x$ and $\sin z$ must then be (strictly) positive, we see that $\sin y$ and $\cos y$ must be, as well.

Substituting $\cos^2y\to 1-\sin^2 y$ into OP's equation $(1)$ gives a quadratic in $\sin y$ we can solve to get
$$\sin y = \frac{ 1 - \cos x }{ 1 + \cos x } \qquad\text{or}\qquad \sin y = -\frac{ 1 + \cos x }{ 1 - \cos x } \tag{1}$$
As $\sin y$ should be strictly positive, we discard the latter option, and we find
$$\cos y = \frac{2 \sqrt{\cos x}}{ 1 + \cos x} \qquad\qquad \frac{1-\sin y}{1+\sin y} = \cos x \tag{2}$$
Consequently, OP's equation $(3)$ becomes
$$1 - \sqrt{\sin z}(1+\sqrt{\cos x}) = \sqrt{\cos x} \quad\to\quad 1 - c - s - s c =  0 \quad\to\quad
s = \frac{1-c}{1+c}\tag{3}$$
where $c := \sqrt{\cos x}$ and $s := \sqrt{\sin z}$. But, then OP's $(2)$ is
$$\frac{(1 - c^2 )^2}{(1 + c^2)^2}(1-s^4) = \frac{8cs^2}{1+c^2} \quad\to\quad 0 = 0 \tag{4}$$
which is to say: OP's equation $(3)$ is not independent of OP's $(1)$ and $(2)$. We have only two equations in three unknowns, hence multiple solutions are not unexpected. Thus,

For $0< c < 1$ we have 
  $$\sin^2 x = 1 - c^4 \qquad \sin y =  \frac{1 - c^2}{1+c^2} \qquad \sin z = \frac{(1-c)^2}{(1+c)^2}$$
$$\cos x= c^2 \qquad \cos y = \frac{2c}{1+c^2} \qquad \cos^2 z =\frac{8 c (1 + c^2)}{(1 + c)^4}$$
  Including $c=0$ and $c=1$ gives solutions $(00)$ and $(0)$, respectively.

A: This is based on modular equations of degree 2. If $k, l$ are elliptic moduli with $l$ of degree 2 over $k$ then we have $$l=\frac{1-k'}{1+k'}\tag{1}$$ where $k'=\sqrt{1-k^2}$.
One should observe that the relation between $\sin x, \sin y$ is such that $\sin y$ is of degree 2 over $\sin x$. This is easily verified by writing $k, l$ in place of $\sin x, \sin y$ in the equation given in question. Doing so we get $$k^2(1-l^2)=4k'l$$ This gives us a quadratic equation in $l$ as $$k^2l^2+4k'l-k^2=0$$ or $$l=\frac{-2k'\pm\sqrt{4k'^2+k^4}}{k^2}$$ Choosing $+$ sign as $l>0$ we get $$l=\frac{2-k^2-2k'}{k^2}=\frac{(1-k')^2}{1-k'^2}=\frac{1-k'}{1+k'}$$ as in equation $(1)$.
Similarly $\sin z$ of degree 2 over $\sin y$ and hence $\sin z$ is of degree 4 over $\sin x$. Let $l_1=\sin z$ so that $$l_1=\frac{1-l'}{1+l'}$$ Next we have $$1+\sqrt{\cos x} =1+\sqrt{k'}=1+\sqrt{\frac{1-l}{1+l}}$$ The equation to be proved  can now be rewritten as $$\sqrt{\sin z} =\dfrac{1-\sqrt{\dfrac{1-l}{1+l}}} {1+\sqrt{\dfrac{1-l} {1+l} } } $$ or $$l_1=\sin z=\frac{2-2\sqrt{1-l^2} } {2+2\sqrt{1-l^2}}=\frac{1-l'}{1+l'} $$ as expected.
The given numerical values for the moduli are taken from a table of singular moduli. In particular the value $$\sin y=(\sqrt{10}-3)(\sqrt{2}-1)^2$$ is $k_{10}$ where $$k_n=\frac{\vartheta_2^2(e^{-\pi\sqrt{n}})}{\vartheta_3^2(e^{-\pi\sqrt{n}})} $$ and $$\vartheta_2(q)=\sum_{r\in\mathbb{Z}}q^{(r+(1/2))^2},\,\vartheta_3(q)=\sum_{r\in\mathbb {Z}} q^{r^2} $$ And therefore the values of $\sin x, \sin z$ are $k_{5/2},k_{40}$ respectively. 
