Question

2sin^2+2sinx*cos2x-1/под конем cosx=0

Answer 1

сosx>0⇒x∈(π/2+2πn;π/2+2πn,n∈z)
2sin²x+2sinx*cosx-sin²x-cos²x=0/cos²x
tg²x+2tgx-1=0
tgx=a
a²+2a-1=0
D=4+4=8
a1=(-2-2√2)/2=-1-√2⇒tgx=-1-√2⇒x=-arctg(1+√2)+2πn,n∈z
a2=-1+√2⇒tgx=√2-1⇒x=arctg(√2-1)+2πk,k∈z