Question

sin^4x – cos^4x + 2cos^2x = 1.

Answer 1

$sin^4x-cos^4x+2cos^2x=1\ cos^2x=t\ 1-2cos^4x+2cos^2x=1\ -2t^2+2t=0\ 2t(1-t)=0\ t=0\ t=1\ cosx=0\ cosx=+-1\ x=2pi*n\ x=2pi*n+/-pi\ x=-frac{pi}{2}+pi*n$
     
Ответ  $x=pi*n$