2sin^2x+cos^2x=5sinx cosx

2sin²x + cos²x = 5sinxcosx 2sin²x - 5sinxcosx + cos²x = 0 2tg²x - 5tgx + 1 = 0 Пусть t = tgx. 2t² - 5t + 1 = 0 D = 25 - 2*4 = 25 - 8 = 17 t₁ = (5 + √17)/4 t₂ = (5 - √17)/4 Обратная замена: tgx = (5 + √17)/4 x = arctg((5 + √17)/4) + πn, n  ∈ Z. tgx = (5 - √17)/4 x = ((5 - √17)/4) + πn, n ∈ Z