5cos²x-6cosx+1=0

И

2ctgx-3tgx+1=0

MoskowCorp MoskowCorp    3   23.11.2020 18:29    3

Ответы
yuliamoisei yuliamoisei  23.12.2020 18:31

5cos²x - 6cosx + 1 = 0,

cosx = а,

5а² - 6а + 1 = 0,

Д = (-6)² - 4*5*1 = 36 - 20 = 16,

а1 = (6 + 4) / 2*5 = 10/10 = 1,

а2 = (6 - 4) / 2*5 = 2/10 = 1/5,

cosx = а1,

cosx = 1,

х1 = 2πn, n ∈ Z,

cosx = а2,

cosx = 1/5,

х2 = ±arccos (1/5) + 2πn, n ∈ Z,

2ctgx - 3tgx + 1 = 0,

2/(tgx) - 3tgx + 1 = 0,   (* tgx)

2tgx - 3tg²x + 1 = 0,

3tg²x - 2tgx - 1 = 0,

tgx = а,

3а² - 2а - 1 = 0,

Д = (-2)² - 4*3*(-1) = 4 + 12 = 16,

а1 = (2 + 4) / 2*3 = 6/6 = 1,

а2 = (2 - 4) / 2*3 = -2/6 = -1/3,

tgx = а1,

tgx = 1,

х = arctg1 + πn, n ∈ Z,

x = π/4 + πn, n ∈ Z,

tgx = а2,

tgx = -1/3,

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