Question

Помогите сроч!!!!!!!!!!!

Answer 1

Сos²x-sin²x-4sinxcosx-2sin²x-2cos²x=0
3sin²x+4sinxcosx+cos²x=0/cos²x
3tg²x+4tgx+1=0
tgx=a
3a²+4a+1=0
D=16-12=4
a1=(-4-2)/6=-1⇒tgx=-1⇒x=-π/4+πk,k∈z
a2=(-4+2)/6=-1/3⇒tgx=-1/3⇒x=-arctg1/3+πk,k∈z
----------------------------------------
x=-5π/4  и x=-π-arctg1/3 приk=-1
x=-9π/4  и x=-π-arctg1/3 при k=-2