Question

cos2x + sinxcosx + cos^x = 1
Срочно!!!

Answer 1

cos2x + sinx*cosx + cos²x = 1

2cos²x - 1 + sinx*cosx + cos²x - 1 = 0

3 cos²x + sinx*cosx - 2 = 0

3 cos²x + sinx*cosx - 2sin²x-2cos²x = 0

cos²x + sinx*cosx - 2sin²x = 0 | : cos²x ≠ 0

1 + tgx - 2tg²x = 0

2tg²x - tgx - 1 = 0

Пусть tgx = t, t ∈ R

2t² - t - 1 = 0

D = 1 +4*2 = 9, √D = 3

t1 = 1 + 3 /4 = 1

t2 = 1 - 3 /4 = - 1/2

tg x = 1 или tgx = -1/2

x = π/4 + πk, k∈Z и x = -arctg 1/2 + πk, k∈ Z