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,
∅
2)26*2=52(р.)-2 календаря
3)52+48=100(р.)
ответ:на всю покупку Ане денег хватит. (хватит)