Question

Sin^2x+sinxcosx=2cos^2x

Answer 1

Поделим на cos²x≠0
tg²x+tgx-2=0
a=tgx
a²+a-2=0
a1+a2=-1 U a1*a2=-2
a1=-2⇒tgx=-2⇒x=-arctg2+πn
a2=1⇒tgx=1⇒x=π/4+πn

Answer 2

""""""""""""""""""""""""""""""""""""""""