Решите : 1) √3 cos3x = sin3x 2) 2 sin( π - 3x) - вопрос №13332232307079106 от ayiina264 19.09.2021 23:41

Question

Алгебра: Решите : 1) √3 cos3x = sin3x 2) 2 sin( π - 3x) + cos( 2π - 3x) = 0

ayiina264 · Answer

√3cos3x = sin3x (cos3x ≠ 0)√3 = tg3x3x = π/3 + πn, n ∈ Zx = π/9 + πn/3, n ∈ Z2sin(π - 3x) + cos(2π - 3x) = 0-2sin(-3x) + cos(3x - 2π) = 02sin3x + cos3x = 02sin3x = -cos3x2tg3x = -1tg3x = -1/23x = arctg(-1/2) + πn, n ∈ Zx = (arctg(-1/2))/3 + πn/3, n ∈ Z....