Question

Решите уравнение:

2(cos^2)x-2sin2x+1=0

Answer 1

2сos²x-4sinxcosx+cos²x+sin²x=0
sin²x-4sinxcosx+3cos²x=0/cos²x≠0
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

Answer 2

2сos²x-4sinxcosx+cos²x+sin²x=0
sin²x-4sinxcosx+3cos²x=0/cos²x≠0
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