Question

sin^2 x - 2cos^2 x=sin2x

Answer 1

sin²x-sinxcosx-2cos²x=0 /cos²x≠0
tg²x-tgx-2=0
tgx=a
a²-a-2=0
a1+a2=1 U a1*a2=-2
a1=-1⇒tgx=-1⇒x=-π/4+πn
a2=2⇒tgx=2⇒x=arctg2+πn