Answer:
Step-by-step explanation:
From what I can come up with with no actual factual information is that the equation is
[tex]\frac{1+sin^4x-cos^4x}{1-sin^6x-cos^6x}=\frac{2cos^2x}{3}[/tex] I'm going with that because it actually solves pretty easily, which is usually the way this goes. Begin by cross multiplying:
[tex]3(1+sin^4x-cos^4x)=2cos^2x(1-sin^6x-cos^6x)[/tex] and distribute:
[tex]3+3sin^4x-3cos^4x=2cos^2x-2cos^2xsin^6x-2cos^8x[/tex] and then break up the powers of sin into squared sines:
[tex]3+3[sin^2x*sin^2x]-3cos^4x=2cos^2x-2cos^2x[sin^2x*sin^2x*sin^2x]-2cos^8x[/tex] and use the Pythagorean identity to change those sines into cosines:
[tex]3+3[(1-cos^2x)(1-cos^2x)]-3cos^4x=2cos^2x-2cos^2x[(1-cos^2c)(1-cos^2x)(1-cos^2x)]-2cos^8x[/tex]
and the FOIL out those terms inside the square brackets:
[tex]3+3[1-2cos^2x+cos^4x]-3cos^4x=2cos^2x-2cos^2x[1-3cos^2x+3cos^4x-cos^6x]-2cos^8x[/tex]
and distribute again on both sides to get rid of the brackets entirely:
[tex]3+3-6cos^2x+3cos^4x-3cos^4x=2cos^2x-2cos^2x+6cos^4x-6cos^6x+2cos^8x-2cos^8x[/tex]
and combining like terms cancels a lot of that out to leave:
[tex]6-6cos^2x=-6cos^6x+6cos^4x[/tex] and get everything on one side to factor:
[tex]0=-6cos^6x+6cos^4x+6cos^2x-6[/tex] and factor out the -6:
[tex]0=-6(cos^6x-cos^4x-cos^2x+1)[/tex] and now we'll factor by grouping, so grouping everything together to get ready for that:
[tex]0=-6[(cos^6x-cos^4x)-(cos^2x+1)][/tex] and begin the factoring:
[tex]0=6[cos^4x(cos^2x-1)-1(cos^2x-1)][/tex] and continuing the factoring:
[tex]0=-6(cos^2x-1)(cos^4x-1)\\0=-6(cos^2x-1)(cos^2x-1)(cos^2x+1)[/tex] and by the Zero Product Property (and we won't list that duplicated factor cuz it's not necessary):
[tex]cos^2x-1=0[/tex] or [tex]cos^2x+1=0[/tex]
Solving the first:
[tex]cos^2x=1[/tex] so
x = 0 + 2kπ and solving the second one:
[tex]cos^2x=-1[/tex] which is not solvable over the real number system.
So x = 0 + 2kπ
and I really hope that's what you were looking for!
