Question

sin^2x+cos^2*2x+cos^2(3пи/2+2x)+2cosx*tgx=1

Answer 1

$sin^2x+cos^22x+cos^2(frac{3pi}{2}+2x)+2cosx*tgx=1 \ sin^2x+cos^22x+sin^22x+2cosx*tgx-cos^2x-sin^2x=0 \ -cos^2x+2cosxtgx+1=0 \ cos^2x-2cosxtgx+1=0 \ tg^2x-2tgx+1=0$

 

d=4-4=0

x=2:2=1

tgx=1

x=p:4 + pk, k принадлежит Z