Solve 2r*cos(x)=rsqrt{4-n^2}

$$2 r \times \cos ( x ) = r \sqrt { 4 - n ^ { 2 } }$$

Solve for n (complex solution)

$$\left\{\begin{matrix}n=-\sqrt{2\left(-\cos(2x)+1\right)}\text{; }n=\sqrt{2\left(-\cos(2x)+1\right)}\text{, }&arg(\cos(x))<\pi \\n=2\text{; }n=-2\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\\n\in \mathrm{C}\text{, }&r=0\end{matrix}\right.$$

Solve for r (complex solution)

$$\left\{\begin{matrix}\\r=0\text{, }&\text{unconditionally}\\r\in \mathrm{C}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }x=-i\ln(\frac{\sqrt{4-n^{2}}-\sqrt{-n^{2}}}{2})+2\pi n_{2}\text{ or }\exists n_{1}\in \mathrm{Z}\text{ : }x=-i\ln(\frac{\sqrt{4-n^{2}}+\sqrt{-n^{2}}}{2})+2\pi n_{1}\end{matrix}\right.$$

Solve for n

$$\left\{\begin{matrix}n=2\sin(x)\text{; }n=-2\sin(x)\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }\left(x\geq 2\pi n_{1}+\frac{3\pi }{2}\text{ and }x\leq 2\pi n_{1}+\frac{5\pi }{2}\right)\\n\in \begin{bmatrix}-2,2\end{bmatrix}\text{, }&r=0\end{matrix}\right.$$

Solve for r

$$\left\{\begin{matrix}r=0\text{, }&|n|\leq 2\\r\in \mathrm{R}\text{, }&\exists n_{1}\in \mathrm{Z}\text{ : }\left(x\geq \frac{\pi \left(4n_{1}+3\right)}{2}\text{ and }x\leq \frac{\pi \left(4n_{1}+5\right)}{2}\right)\text{ and }|n|=2|\sin(x)|\text{ and }|n|\leq 2\end{matrix}\right.$$