[University Mathematics:Calculus 1] Is this the only solution to this question or are there other ways to solve it?

This comment was posted to reddit on Dec 04, 2022 at 1:17 pm and was deleted within 13 minutes.

Not the obvious solution, but: cos^{4(x)*(2cos^2(x)} - 3) + sin^{4(x)*(2sin^2(x)} - 3)=-1

2cos^2(x) - 1 = cos2x and 1 - 2sin^2(x) = cos2x then:

cos^4(x)*(cos2x - 2) + sin^4(x)*(-cos2x - 2)=-1

cos^4(x)*cos2x - 2cos^4(x) - sin^{4^x*cos2x} \ -2sin^4(x) = -1

cos2x(cos^{4(x)-sin^4(x))} -2(sin^{4(x)+cos^4(x))} =-1

cos2x(cos^{2(x)-sin^2(x))}(cos^{2(x)+sin^2(x))} \ -2(sin^4(x)+2sin^{2(x)*cos^{2(x)+cos^4(x)}} \ -2sin^2(x)cos^2(x)) = -1

cos^2(x) + sin^2(x) = 1 and cos^2(x) - sin^2(x) = \ cos2x and sin^4(x) + 2sin^{2(x)*cos^2(x)} +cos^4(x)=\ (sin^{2(x)+cos^2(x))²} = 1 then:

cos2xcos2x - 2(1 - 2sin^2(x)cos^2(x)) = -1

2sinxcosx = sin2x => 2sin^{2(x)*cos^2(x)} = \ 0,5*sin^2(2x) then:

cos^2(2x) - 2*(1 - 0,5sin^2(2x)) = -1

cos^2(2x) -2 + sin^2(2x) = -1

1 - 2 = -1 -1 = -1