Question

sin^2x+sinxsin(3п/2-x)-cos2x=1

Answer 1

Sin^2x+sinxsin(3п/2-x)-cos2x=1
sin²x-sinxcosx-cos²x+sin²x-sin²x-cos²x=0
sin²x-sinxcosx-2cos²x=0/cos²x
tg²x-tgx-2=0
tgx=a
a²-a-2=0
a1+a2=1 U a1*a2=-2
a1=-1⇒tgx=-1⇒x=-π/4+πk,k∈z
a2=2⇒tgx=2⇒x=arctg2+πk,k∈z

Answer 2

Cos^2x-sinxcosx=sin^2x+cos^2x
sin^2x+sinxcosx=0
sinx(sinx+cosx)=0
sinx=0
x=Пk k=-3 x=-3П
k=-2 x=-2П
sinx=-cosx
tgx=-1
x=-П/4+Пk
k=-2
x=-9П/4