1+sin^4-cos^4a/1-sin^6-cos^6a=2/3co s^2a

Step-by-step explanation:

From what I can come up with with no actual factual information is that the equation is

I'm going with that because it actually solves pretty easily, which is usually the way this goes. Begin by cross multiplying:

and distribute:

and then break up the powers of sin into squared sines:

 and use the Pythagorean identity to change those sines into cosines:

and the FOIL out those terms inside the square brackets:

and distribute again on both sides to get rid of the brackets entirely:

and combining like terms cancels a lot of that out to leave:

and get everything on one side to factor:

and factor out the -6:

and now we'll factor by grouping, so grouping everything together to get ready for that:

and begin the factoring:

and continuing the factoring:

and by the Zero Product Property (and we won't list that duplicated factor cuz it's not necessary):

    or      

Solving the first:

so

x = 0 + 2kπ  and solving the second one:

which is not solvable over the real number system.

So x = 0 + 2kπ

and I really hope that's what you were looking for!