Question

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

Answer 1

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