Question

5cos²x-2sinxcosx+sin²x=4

Answer 1

5cos²x-2sinxcosx+sin²x-4sin²x-4cos²x=0
3sin²x+2sinxcosx-cos²x=0  /cos²x≠0
3tg²x+2tgx-1=0
tgx=a
3a²+2a-1=0
D=4+12=16
a1=(-2-4)/6=-1⇒tgx=-1⇒x=-π/4+πn
a2=(-2+4)/6=1/3⇒tgx=1/3⇒x=arctg1/3+πn