Solve Cd^-1[frac{sqrt{1+sinn}+sqrt{1-sinn}}{sqrt{1+sinn}-sqrt{1-sinn}}]=n/2

$$C d ^ { - 1 } [ \frac { \sqrt { 1 + \sin n } + \sqrt { 1 - \sin n } } { \sqrt { 1 + \sin n } - \sqrt { 1 - \sin n } } ] = \frac { n } { 2 }$$

Solve for C

$$C=-\frac{dn\left(\sqrt{-\left(\sin(n)\right)^{2}+1}-1\right)}{2\sin(n)}$$

$$d\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }n=\pi n_{1}$$

Steps for Solving Linear Equation

The equation is in standard form.

$$\frac{\sqrt{\sin(n)+1}+\sqrt{-\sin(n)+1}}{d\left(\sqrt{\sin(n)+1}-\sqrt{-\sin(n)+1}\right)}C=\frac{n}{2}$$

Divide both sides by d^{-1}\left(\sqrt{1+\sin(n)}+\sqrt{1-\sin(n)}\right)\left(\sqrt{1+\sin(n)}-\sqrt{1-\sin(n)}\right)^{-1}.

$$\frac{\frac{\sqrt{\sin(n)+1}+\sqrt{-\sin(n)+1}}{d\left(\sqrt{\sin(n)+1}-\sqrt{-\sin(n)+1}\right)}Cd\left(\sqrt{\sin(n)+1}-\sqrt{-\sin(n)+1}\right)}{\sqrt{\sin(n)+1}+\sqrt{-\sin(n)+1}}=\frac{n}{2\times \frac{\sqrt{\sin(n)+1}+\sqrt{-\sin(n)+1}}{d\left(\sqrt{\sin(n)+1}-\sqrt{-\sin(n)+1}\right)}}$$

Dividing by d^{-1}\left(\sqrt{1+\sin(n)}+\sqrt{1-\sin(n)}\right)\left(\sqrt{1+\sin(n)}-\sqrt{1-\sin(n)}\right)^{-1} undoes the multiplication by d^{-1}\left(\sqrt{1+\sin(n)}+\sqrt{1-\sin(n)}\right)\left(\sqrt{1+\sin(n)}-\sqrt{1-\sin(n)}\right)^{-1}.

$$C=\frac{n}{2\times \frac{\sqrt{\sin(n)+1}+\sqrt{-\sin(n)+1}}{d\left(\sqrt{\sin(n)+1}-\sqrt{-\sin(n)+1}\right)}}$$

Divide \frac{n}{2} by d^{-1}\left(\sqrt{1+\sin(n)}+\sqrt{1-\sin(n)}\right)\left(\sqrt{1+\sin(n)}-\sqrt{1-\sin(n)}\right)^{-1}.

$$C=\frac{dn\left(-|\cos(n)|+1\right)}{2\sin(n)}$$

Solve for d

$$d=\frac{2C\left(\sqrt{-\left(\sin(n)\right)^{2}+1}+1\right)}{n\sin(n)}$$

$$\nexists n_{1}\in \mathrm{Z}\text{ : }n=\pi n_{1}\text{ and }C\neq 0$$