Question

cos 2x+sin2x+2cos^2x=0

Answer 1

cos^2x-sin^2x+2sinxcosx+2cos^2x=0|:cos^2x
1-tg^2x+2tgx+2=0
-tg^2x+2tgx+3=0 tgx=t
-t^2+2t+3=0
D=4-4*(-1)*3=16
t1=-2+4/-2=-1
t2=-2-4/-2=3
tgx=-1
x=-п/4+пn,n€z
tgx=3
x=arctg3+пn,n€z

Answer 2

спасибо