Q:

Solve system of non-linear equations

I am trying to solve the following system of four equations. I have tried using the "rootSolve" package but it does not seem like I can find a solution this way. 
The code I am using is the following:
model <- function(x) {
F1 <- sqrt(x[1]^2 + x[3]^2) -1
F2 <- sqrt(x[2]^2 + x[4]^2) -1
F3 <- x[1]*x[2] + x[3]*x[4]
F4 <- -0.58*x[2] - 0.19*x[3]
c(F1 = F1, F2 = F2, F3 = F3, F4 = F4)
}
(ss <- multiroot(f = model, start = c(0,0,0,0)))

But it gives me the following error:
Warning messages:
1: In stode(y, times, func, parms = parms, ...) :
error during factorisation of matrix (dgefa);         singular matrix
2: In stode(y, times, func, parms = parms, ...) : steady-state not reached

I have changed the starting values, as suggested in another similar answer, and for some I can find a solution. 
However, this system - according to the source I am using - should have an uniquely  identified solution. 
Any idea about how to solve this system?
Thanks you!

A:

Your system of equations has multiple solutions.
I use a different package to solve your system: nleqslv as follows:
library(nleqslv)

model <- function(x) {
   F1 <- sqrt(x[1]^2 + x[3]^2) - 1
   F2 <- sqrt(x[2]^2 + x[4]^2) - 1
   F3 <- x[1]*x[2] + x[3]*x[4]
   F4 <- -0.58*x[2] - 0.19*x[3]
   c(F1 = F1, F2 = F2, F3 = F3, F4 = F4)
}

#find solution
xstart  <-  c(1.5, 0, 0.5, 0)
nleqslv(xstart,model)

This gets the same solution as the answer of Prem.
Your system however has multiple solutions.
Package nleqslv provides a function to search for solutions given a matrix of different starting values. You can use this
set.seed(13)
xstart <- matrix(runif(400,0,2),ncol=4)
searchZeros(xstart,model)

(Note: different seeds may not find all four solutions)
You will see that there are four different solutions:
$x
     [,1]          [,2]          [,3] [,4]
[1,]   -1 -1.869055e-10  5.705536e-10   -1
[2,]   -1  4.992198e-13 -1.523934e-12    1
[3,]    1 -1.691309e-10  5.162942e-10   -1
[4,]    1  1.791944e-09 -5.470144e-09    1
.......

This clearly suggests that the exact solutions are as given in the following matrix 
xsol <- matrix(c(1,0,0,1,
                 1,0,0,-1,
                -1,0,0,1,
                -1,0,0,-1),byrow=TRUE,ncol=4)

And then do
model(xsol[1,])
model(xsol[2,])
model(xsol[3,])
model(xsol[4,])

Confirmed!
I have not tried to find these solutions analytically but you can see that if x[2] and x[3] are zero then F3 and F4 are zero. The solutions for x[1] and x[4] can  then be immediately found.

