How many complex solutions and real solutions does the polynomial system has? The following polynomial equation has six variables $sb,sc,sd,cb,cc,cd$ and 7 equations
$$\begin{cases}
    &-sb-sc=0 \\
    &sb+sb*cc-sc*cb+sb*cd-sd*cb=0 \\
    &sc+sc*cb-sb*cc+sc*cd-sd*cc=0 \\
    &sd+sd*cb-sb*cd+sd*cc-sc*cd=0 \\
    &sb^2+cb^2=1 \\
     &sc^2+cc^2=1 \\
      &sd^2+cd^2=1 \\
\end{cases}$$
I am wondering how many complex solutions and real solutions does it have? Are they finite or infinite?
I tried numerical irreducible decomposition in complex space and the result shows that it has 3 dim-1 components and 5 dim-0 component.
Dimension 1: 3 classified components
-----------------------------------------------------
   degree 2: 3 components

Dimension 0: 5 classified components
-----------------------------------------------------
   degree 1: 5 components

But I have no idea about real solutions, how to determine how many real solutions are there and how to determine them?
Update on the context:
The system is derived from the following system by fixing $a=0$. In order to solve it, I transfer it to the polynomial system.
$$\begin{cases}
&sin(a-b)+sin(a-c)+sin(a-d)=0 \\
&sin(b-a)+sin(b-c)+sin(b-d)=0 \\
&sin(c-a)+sin(c-b)+sin(c-d)=0 \\
&sin(d-a)+sin(d-b)+sin(d-c)=0
\end{cases}$$
 A: (Partial answer, about the real solutions.)
Let $\,A=\cos a + i \sin a\,$ and similar for $\,B,C,D\,$, then $\,|A|=|B|=|C|=|D|=1\,$, and:
$$
\text{Im}\left(\frac{A+B+C+D}{A}\right)=\text{Im}\left(\frac{B+C+D}{A}\right)=\sin(b-a)+\sin(c-a)+\sin(d-a)=0
$$

*

*If $\,S = A+B+C+D \ne 0\,$ this is equivalent to $\,A\,$ being a real multiple of $\,S\,$, and the same goes for $\,B,C,D\,$, so $A = \alpha\,S$, $\;B = \beta\,S$, $\;C = \gamma\,S$, $D = \delta\,S\;$ for some $\,\alpha,\beta,\gamma,\delta \in \mathbb R\,$.
Adding the equations gives $\,S = (\alpha+\beta+\gamma+\delta) \,S$ $\implies S=0 \;\lor\; \alpha+\beta+\gamma+\delta = 1$.
$S=0\,$ gives $\,A=\alpha\,S=0\,$ which violates $\,|A|=1\,$, otherwise $\,A = \alpha\,S$ $\implies |\alpha| \, |S| = |A| = 1$ $\implies \alpha = \pm \frac{1}{|S|}\,$. Then, the sum condition requires that $\,\pm \frac{1}{|S|}\pm \frac{1}{|S|}\pm \frac{1}{|S|}\pm \frac{1}{|S|}=1\,$ for some choice of signs. This can only happen in two cases (not counting permutations of the variables):

*

*$\,\alpha=\beta=\gamma=\delta=\frac{1}{|S|}\,$ and $\,|S|=4\,$ $\implies A=B=C=D$ $\implies \color{blue}{a\equiv b\equiv c\equiv d}$ $\pmod{2\pi}\,$, which is easily verified to be a solution for any $\,a \in \mathbb R\,$;


*$\,-\alpha=\beta=\gamma=\delta=\frac{1}{|S|}\,$ and $\,|S|=2\,$ $\implies -A=B=C=D$ $\,\implies \color{blue}{a+\pi\equiv b\equiv c\equiv d}\,$ $\pmod{2\pi}\,$, which is also easily verified to be a solution for any $\,a \in \mathbb R\,$.




*This leaves the case $\,S=0\,$, when $\,\text{Im}\left(\frac{B+C+D}{A}\right)=\text{Im}\left(\frac{-A}{A}\right)=0\,$, so the equations are satisfied automatically. What remains to be determined is the solution set of the equation $\,S=A+B+C+D=0\,$ subject to the constraint $\,|A|=|B|=|C|=|D|=1\,$.
For that, it's enough to note that four unit vectors which add to $\,0\,$ form a rhombus when placed head-to-tail, so they must be pairwise opposite, or (again not counting permutations) $-A=-B=C=D$ $\implies \color{blue}{a+\pi\equiv b+\pi\equiv c\equiv d}\,$ $\pmod{2\pi}\,$.
Putting it $\color{blue}{\text{all}}$ together, the general solution in reals is $a\equiv b\equiv c\equiv d \pmod{\pi}$.
