2cos^x-sinxcosx+5sin^x=3

Question

2cos^x-sinxcosx+5sin^x=3

Answer 1

2cos²x-sinxcosx+5sin²x=3cos²x+3sin²x(путем замены cos²x+sin²x= 1 ) 
2sin²x-sinxcosx-cos²x=0 
2tg²x-tgx-1=0 
p=tgx 
2p²-p-1=0 
p₁=1 
p₂=-½ 
tgx=1; x₁=π/4+πN 
tgx=-½; x₂=-arctg(½)+πK