Question

3sin^2x-4sinx*cosx+5cos^2x=2
Решите уравнение

Answer 1

3sin²x-4sinxcosx+5cos²x-2sin²x-2cos²x=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