Решите уравнение:

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

2сos²x-4sinxcosx+cos²x+sin²x=0sin²x-4sinxcosx+3cos²x=0/cos²x≠0tg²x-4tgx+3=0tgx=aa²-4a+3=0a1+a2=4 U a1*a2=3a1=1⇒tgx=1⇒x=π/4+πn,n∈Za2=3⇒tgx=3⇒x=arctg3+πk,k∈Z

2сos²x-4sinxcosx+cos²x+sin²x=0sin²x-4sinxcosx+3cos²x=0/cos²x≠0tg²x-4tgx+3=0tgx=aa²-4a+3=0a1+a2=4 U a1*a2=3a1=1⇒tgx=1⇒x=π/4+πn,n∈Za2=3⇒tgx=3⇒x=arctg3+πk,k∈Z