Question

Sin 2x +4sin^2x=2cos^2x

Answer 1

4sin²x+2sinxcosx-2cos²x=0/2cos²x
2tg²x+tgx-1=0
tgx=a
2a²+a-1=0
D=1+8=9
a1=(-1-3)/4=-1⇒tgx=-1⇒x=-π/4+πn,n∈z
a2=(-1+3)/4=1/2⇒tgx=1/2⇒x=arctg0,5+πk,k∈z