Question

3sin(в квадрате)x-4sinxcosx+5cos(в квадрате)x=2

Answer 1

3sin²x-4sinxcosx+5cos²x-2sin²x-2cos²x=0
sin²x-4sinxcosx+3cos²x=0/cos²x
tg²x-4tgx+3=0
tgx=a
a²-4a+3=0
a1+a2=4 U a1*a2=3
a1=1⇒tgx=1⇒x=π/4+πn,n∈z
a2=3⇒tgx=3⇒x=arctg3+πk,k∈z