nonlinear system of equations

Question

Could you help me to solve the following system of $3$ equations and $3$ variables?

$$\left\{\begin{array} {lllll} 2x+y+z=0\\ x-z-\sin y=0\\ x-y+\sin z=0\\ \end{array}\right. $$

Answer 1

From the first equation, you can eliminate $x=-\frac{1}{2} (y+z)$; so now the second equation becomes $$y+2 \sin (y)+3 z=0$$ from which $z=-\frac{1}{3} (y+2 \sin (y))$. Replacing in the third, $$4 y-\sin (y)+3 \sin \left(\frac{1}{3} (y+2 \sin (y))\right)=0$$ for which I doubt that you could find an explicit solution except $??$ (what you could have guessed by inspection).

I am sure that you can take from here.

Edit

The problem would have been much more interesting using as third equation $$x-y+\cos(z)=0$$ Try to play with it; it is interesting. You just need to replace $3\sin(.)$ by $3\cos(.)$

Answer 2

There is one fairly simple solution to start with.

Use the first equation to eliminate $x$ from the other two equations.
Now the second equation gives you $z$ as a function of $y$. Plug that into the third equation, and you get a single equation for $y$.
Plot this equation using Wolfram Alpha, or your calculator, to estimate the solutions.