1) 12sin5x=cos10x+7 2) 3tg²x -8cos²x+1=0 3) 3sinx+cosx=1 - вопрос №1125107238103311347503 от sport22 21.08.2022 03:48

Question

Алгебра: 1) 12sin5x=cos10x+7 2) 3tg²x -8cos²x+1=0 3) 3sinx+cosx=1

sport22 · Answer

1) 12Sin5x = Cos10x + 712Sin5x - Cos10x - 7 = 012Sin5x - (1 - 2Sin²5x) - 7 = 012Sin5x - 1 + 2Sin²5x - 7 = 02Sin²5x + 12Sin5x - 8 = 0Sin²5x + 6Sin5x - 4 = 0Sin5x = m ,  - 1 ≤ m ≤ 1m² + 6m - 4 = 0D = 6² - 4 * (- 4) = 36 + 16 = 52 = (2√13)²Решений нет...