Differential equation sec^2x*tgy*dy+sec^2ytgxdx=0

Given the equation:
$$\tan{\left(x \right)} \sec^{2}{\left(y{\left(x \right)} \right)} + \tan{\left(y{\left(x \right)} \right)} \sec^{2}{\left(x \right)} \frac{d}{d x} y{\left(x \right)} = 0$$
This differential equation has the form:

f1(x)*g1(y)*y' = f2(x)*g2(y),

where
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(y \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - \sin{\left(2 x \right)}$$
$$\operatorname{g_{2}}{\left(y \right)} = \frac{1}{\sin{\left(2 y{\left(x \right)} \right)}}$$
We give the equation to the form:

g1(y)/g2(y)*y'= f2(x)/f1(x).

Divide both parts of the equation by g2(y)
$$\frac{1}{\sin{\left(2 y{\left(x \right)} \right)}}$$
we get
$$\sin{\left(2 y{\left(x \right)} \right)} \frac{d}{d x} y{\left(x \right)} = - \sin{\left(2 x \right)}$$
We separated the variables x and y.

Now, multiply the both equation sides by dx,
then the equation will be the
$$dx \sin{\left(2 y{\left(x \right)} \right)} \frac{d}{d x} y{\left(x \right)} = - dx \sin{\left(2 x \right)}$$
or
$$dy \sin{\left(2 y{\left(x \right)} \right)} = - dx \sin{\left(2 x \right)}$$

Take the integrals from the both equation sides:
- the integral of the left side by y,
- the integral of the right side by x.
$$\int \sin{\left(2 y \right)}\, dy = \int \left(- \sin{\left(2 x \right)}\right)\, dx$$
Detailed solution of the integral with y
Detailed solution of the integral with x
Take this integrals
$$- \frac{\cos{\left(2 y \right)}}{2} = Const + \frac{\cos{\left(2 x \right)}}{2}$$
Detailed solution of the equation
We get the simple equation with the unknown variable y.
(Const - it is a constant)

Solution is:
$$\operatorname{y_{1}} = y{\left(x \right)} = \pi - \frac{\operatorname{acos}{\left(C_{1} - \cos{\left(2 x \right)} \right)}}{2}$$
$$\operatorname{y_{2}} = y{\left(x \right)} = \frac{\operatorname{acos}{\left(C_{1} - \cos{\left(2 x \right)} \right)}}{2}$$