Solve sec^2(y)dy=cos(x)dx

$$\sec ^ { 2 } ( y ) d y = \cos ( x ) d x$$

Solve for d (complex solution)

$$\left\{\begin{matrix}d=0\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }y=\pi n_{1}+\frac{\pi }{2}\\d\in \mathrm{C}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }y=\frac{\pi \left(2n_{1}+1\right)}{2}\text{ and }x\left(-\frac{1}{e^{ix+2iy}}-\frac{1}{e^{2iy-ix}}-e^{ix+2iy}-e^{2iy-ix}-4\cos(x)\right)=-8y\end{matrix}\right.$$

Solve for d

$$\left\{\begin{matrix}d=0\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }y=\pi n_{1}+\frac{\pi }{2}\\d\in \mathrm{R}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }y=\frac{\pi \left(2n_{1}+1\right)}{2}\text{ and }x\cos(x)=\frac{y}{\left(\cos(y)\right)^{2}}\end{matrix}\right.$$