Zgjidh sinx+cosy=4

$$\sin x + \cos y = 4$$

Gjej x

$$\left\{\begin{matrix}x=-i\ln(\frac{-2i\cos(y)-\sqrt{2}\sqrt{16\cos(y)-\cos(2y)-31}+8i}{2})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&\frac{-2i\cos(y)-\sqrt{2}\sqrt{16\cos(y)-\cos(2y)-31}+8i}{2}\neq 0\\x=-i\ln(\frac{-2i\cos(y)+\sqrt{2}\sqrt{16\cos(y)-\cos(2y)-31}+8i}{2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&\frac{-2i\cos(y)+\sqrt{2}\sqrt{16\cos(y)-\cos(2y)-31}+8i}{2}\neq 0\end{matrix}\right.$$

Gjej y

$$\left\{\begin{matrix}y=-i\ln(\frac{-2\sin(x)-\sqrt{2}\sqrt{-\cos(2x)-16\sin(x)+31}+8}{2})+2\pi n_{1}\text{, }n_{1}\in \mathrm{Z}\text{, }&\frac{-2\sin(x)-\sqrt{2}\sqrt{-\cos(2x)-16\sin(x)+31}+8}{2}\neq 0\\y=-i\ln(\frac{-2\sin(x)+\sqrt{2}\sqrt{-\cos(2x)-16\sin(x)+31}+8}{2})+2\pi n_{2}\text{, }n_{2}\in \mathrm{Z}\text{, }&\frac{-2\sin(x)+\sqrt{2}\sqrt{-\cos(2x)-16\sin(x)+31}+8}{2}\neq 0\end{matrix}\right.$$