How would I solve this non linear systems of equation with trig.

Question

How would I solve this. I'm kinda stuck... $$\cos(y) + \cos(z) = -1$$ $$\sin(y) + \sin(z) = 0$$

Answer 1

I would square the equations and add them and subtract them. You get some expressions for $\cos(y\pm z)$. Just remember to check the result with the original equations, since squaring them will transform $-1$ into $+1$

Answer 2

By the sum to product identities, $$\cos(y)+\cos(z) = 2\cos(\frac{y+z}2)\cos(\frac{y-z}2)=-1$$$$\sin(y)+\sin(z) = 2\sin(\frac{y+z}2)\cos(\frac{y-z}2)=0$$

Since the first product isn't zero, neither of the multiplicands is zero. Hence, $\cos(\frac{y-z}2)\neq0\to\sin(\frac{y+z}2)=0\to y+z=2\pi k$ for some $k$. WLOG assume $y+z=2\pi$. So, $2\cos(\frac{y+z}2)=-2$, so $\cos(\frac{y-z}2)=\frac12$. Hence, $\frac{y-z}2=\pm\frac\pi3\to y-z=\pm\frac{2\pi}3$. You now have two equations, two variables. Can you solve from here?